Trigonometry

1. Trigonometry

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One of the sides has length 11 units and another side has length 7 units, as labelled. One of the angles is labelled x.

Solve for x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x= °
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the adjacent side and the hypotenuse. So the trigonometric ratio to use is cos. We can write an equation with the given information and x:

cosθ=adjacenthypotenusecosx=711

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. We need to use the inverse cosine operation: that is how we get the x out of the cosine calculation. As always, we must do the same thing to both sides of the equation.

cosx=711cos1(cosx)=cos1(711)x=cos1(711)=50,47880...50,48°

The missing angle, x, is 50,48°.


Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: tan11°.

INSTRUCTION: Round your answer to two decimal places.
Answer: tan11°
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate tan11°. If you are not sure about how to do that, refer to the manual of your calculator.


STEP: Use your calculator to evaluate tan11° and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

tan11°=0,19438...0,19

The correct answer is 0,19.


Submit your answer as:

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One angle is labelled 54° and one side is 8 units long. Another one of the sides is labelled x.

Solve for x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite and adjacent sides. So the trigonometric ratio to use is tan. We can write an equation with the given information and x:

tanθ=oppositeadjacenttan54°=x8

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. We need to multiply the 8 across to the left side to isolate x. Then evaluate the left side.

tan54°=x8(8)tan54°=x(8)(1,37638...)=x11,01105...=x11,01x

Therefore x is 11,01.


Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows an isosceles triangle. Side MP¯=6,1 is labelled and M^=62,7°. Point Q is on side MN¯ directly across from point P, such that PQ¯ makes a right angle with MN¯.

  1. Which of the following statements must be true about the figure above?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that isosceles triangles are symmetric.


    STEP: Consider the choices and select the correct option
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to select the true statement from the choices in the list. Note that two of the options must be wrong straightaway: ΔMNP is neither a right-angled triangle nor equilateral. From the remaining options, the correct choice is based on the fact that the segment PQ¯ divides ΔMNP into two identical shapes: the segment PQ¯ is a line of symmetry because the triangle is isosceles. Therefore, point Q divides MN¯ into two equal parts.

    The correct choice from the list is: MQ¯ = QN¯.


    Submit your answer as:
  2. Next, find the length of MN¯. Round your answer to one decimal place.

    Answer: The length of MN¯ is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the segment PQ¯ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.


    STEP: Draw a line to create right-angled triangles
    [−1 point ⇒ 2 / 3 points left]

    The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

    The line segment QP¯ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:


    STEP: Use trigonometry to find useful information
    [−1 point ⇒ 1 / 3 points left]

    In ΔMQP (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

    With the information given, we can find both segments MQ¯ and PQ¯. However, for this question only MQ¯ is useful: we want the length of MN¯, which is twice as long as MQ¯. In ΔMQP, the hypotenuse is MP¯=6,1, and the side we want is adjacent to the angle given. Therefore we need to use the cosine ratio. Set up the equation and then solve for the length of MQ¯.

    cosθ=adjacenthypotenusecos(62,7°)=MQ¯6,1(6,1)cos62,7°=MQ¯(6,1)(0,4586...)=MQ¯2,7977...=MQ¯


    STEP: Calculate the final answer
    [−1 point ⇒ 0 / 3 points left]

    We want the length of MN¯, so multiply by two since Q is the mid-point of segment MN¯:

    MN¯=2MQ¯=2(2,7977...)=5,5955...5,6

    Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

    The final answer is: MN¯=5,6.


    Submit your answer as:

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One angle is labelled 52° and one side is 8 units long. Another one of the sides is labelled x.

Solve for x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite side and the hypotenuse. So the trigonometric ratio to use is sin. We can write an equation with the given information and x:

sinθ=oppositehypotenusesin52°=8x

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Rearrange the equation to make x the subject. Then evaluate the right side.

sin52°=8xx=8sin52°=80,78801...=10,15214...10,15

Therefore x is 10,15.


Submit your answer as:

Exercises

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One angle is labelled 48° and one side is 6 units long. Another one of the sides is labelled x.

Solve for x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the adjacent side and the hypotenuse. So the trigonometric ratio to use is cos. We can write an equation with the given information and x:

cosθ=adjacenthypotenusecos48°=x6

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Multiply by 6 to isolate x. Then evaluate the left side of the equation.

cos48°=x6(6)cos48°=x(6)(0,66913...)=x4,01478...=x4,01x

The missing side, x, is 4,01.


Submit your answer as:

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One angle is labelled 56° and one side is 7 units long. Another one of the sides is labelled x.

Solve for x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the adjacent side and the hypotenuse. So the trigonometric ratio to use is cos. We can write an equation with the given information and x:

cosθ=adjacenthypotenusecos56°=x7

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Multiply by 7 to isolate x. Then evaluate the left side of the equation.

cos56°=x7(7)cos56°=x(7)(0,55919...)=x3,91435...=x3,91x

The missing side, x, is 3,91.


Submit your answer as:

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One of the sides has length 15 units and another side has length 8 units, as labelled. One of the angles is labelled x.

Solve for x.

INSTRUCTION: Give your answer rounded to two decimal places.
Answer: x= °
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the adjacent side and the hypotenuse. So the trigonometric ratio to use is cos. We can write an equation with the given information and x:

cosθ=adjacenthypotenusecosx=815

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. We need to use the inverse cosine operation: that is how we get the x out of the cosine calculation. As always, we must do the same thing to both sides of the equation.

cosx=815cos1(cosx)=cos1(815)x=cos1(815)=57,76904...57,77°

The missing angle, x, is 57,77°.


Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: sin38°.

INSTRUCTION: Round your answer to two decimal places.
Answer: sin38°
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate sin38°. If you are not sure about how to do that, refer to the manual of your calculator.


STEP: Use your calculator to evaluate sin38° and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

sin38°=0,61566...0,62

The correct answer is 0,62.


Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: tan52°.

INSTRUCTION: Round your answer to two decimal places.
Answer: tan52°
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate tan52°. If you are not sure about how to do that, refer to the manual of your calculator.


STEP: Use your calculator to evaluate tan52° and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

tan52°=1,27994...1,28

The correct answer is 1,28.


Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: sin5°.

INSTRUCTION: Round your answer to two decimal places.
Answer: sin5°
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate sin5°. If you are not sure about how to do that, refer to the manual of your calculator.


STEP: Use your calculator to evaluate sin5° and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

sin5°=0,08715...0,09

The correct answer is 0,09.


Submit your answer as:

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One of the sides is 7 units long and another side is 6 units long, as labelled. One of the angles is labelled x.

Calculate x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x= °
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite and adjacent sides. So the trigonometric ratio to use is tan. We can write an equation with the given information and x:

tanθ=oppositeadjacenttanx=76

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. We need to use the inverse tangent operation: that is how we get the x out of the tangent calculation. As always, we must do the same thing to both sides of the equation.

tanx=76tan1(tanx)=tan1(76)x=tan1(76)=49,39870...49,4°

Therefore x is 49,4°.


Submit your answer as:

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One angle is labelled 51° and one side is 5 units long. Another one of the sides is labelled x.

Determine the value of x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite and adjacent sides. So the trigonometric ratio to use is tan. We can write an equation with the given information and x:

tanθ=oppositeadjacenttan51°=5x

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Start by rearranging the equation to make x the subject. That means tan51° will end up in the denominator on the right side.

tan51°=5xx=5tan51°=51,23489...=4,04892...4,05

Therefore x is 4,05.


Submit your answer as:

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One angle is labelled 42° and one side is 10 units long. Another one of the sides is labelled x.

Calculate x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite and adjacent sides. So the trigonometric ratio to use is tan. We can write an equation with the given information and x:

tanθ=oppositeadjacenttan42°=10x

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Start by rearranging the equation to make x the subject. That means tan42° will end up in the denominator on the right side.

tan42°=10xx=10tan42°=100,90040...=11,10612...11,11

Therefore x is 11,11.


Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows an isosceles triangle. Side XZ¯=3,2 is labelled and X^=30,8°. Point P is on side XY¯ directly across from point Z, such that ZP¯ makes a right angle with XY¯.

  1. Which of the following statements must be true about the figure above?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that isosceles triangles are symmetric.


    STEP: Consider the choices and select the correct option
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to select the true statement from the choices in the list. Note that two of the options must be wrong straightaway: ΔXYZ is neither a right-angled triangle nor equilateral. From the remaining options, the correct choice is based on the fact that the segment ZP¯ divides ΔXYZ into two identical shapes: the segment ZP¯ is a line of symmetry because the triangle is isosceles. Therefore, point P divides XY¯ into two equal parts.

    The correct choice from the list is: XP¯ = PY¯.


    Submit your answer as:
  2. Now determine the length of XY¯. Round your answer to one decimal place.

    Answer: The length of XY¯ is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the segment ZP¯ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.


    STEP: Draw a line to create right-angled triangles
    [−1 point ⇒ 2 / 3 points left]

    The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

    The line segment PZ¯ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:


    STEP: Use trigonometry to find useful information
    [−1 point ⇒ 1 / 3 points left]

    In ΔXPZ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

    With the information given, we can find both segments XP¯ and ZP¯. However, for this question only XP¯ is useful: we want the length of XY¯, which is twice as long as XP¯. In ΔXPZ, the hypotenuse is XZ¯=3,2, and the side we want is adjacent to the angle given. Therefore we need to use the cosine ratio. Set up the equation and then solve for the length of XP¯.

    cosθ=adjacenthypotenusecos(30,8°)=XP¯3,2(3,2)cos30,8°=XP¯(3,2)(0,8589...)=XP¯2,7486...=XP¯


    STEP: Calculate the final answer
    [−1 point ⇒ 0 / 3 points left]

    We want the length of XY¯, so multiply by two since P is the mid-point of segment XY¯:

    XY¯=2XP¯=2(2,7486...)=5,4973...5,5

    Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

    The final answer is: XY¯=5,5.


    Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows a scalene triangle. Two sides and an angle are given: AC¯=8,2, AB¯=9,7 and A^=58,2°. Point D is on side AB¯ as labelled, such that CD¯ makes a right angle with AB¯.

  1. Which of the following statements must be true about the figure above?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Imagine walking from point A to D and then walking on from D to B. How does this compare to walking straight from A to B without stopping?


    STEP: Consider the choices and select the correct option
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to select the true statement from the choices in the list. The correct option is about the sum of the segments along side AB¯. Point D breaks AB¯ up into two pieces. Therefore we know that together they make the total length of AB¯. Note that we do not know the exact position of point D, only that it is somewhere between A and B.

    The correct choice from the list is: AD¯+DB¯=AB¯.


    Submit your answer as:
  2. Now determine the measure of B^. Round your answer to one decimal place.

    Answer: The size of B^ is °.
    numeric
    HINT: <no title>
    [−0 points ⇒ 7 / 7 points left]

    Use the segment CD¯ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.


    STEP: Draw a line to create right-angled triangles
    [−1 point ⇒ 6 / 7 points left]

    The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

    The line segment DC¯ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:


    STEP: Use trigonometry to find useful information
    [−4 points ⇒ 2 / 7 points left]

    In ΔADC (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

    With the information given, we can find both segments AD¯ and CD¯. Remember that we want to get the measure of B^, and for that we need both of these lengths. Start by calculating the length of AD¯, which will allow us to find DB¯ (because we know that AB¯=9,7). This calculation involves the hypotenuse and the side adjacent to A^, so use the cosine ratio.

    cosθ=adjacenthypotenusecos(58,2°)=AD¯8,2(8,2)cos58,2°=AD¯(8,2)(0,5269...)=AD¯4,3210...=AD¯SinceAD¯+DB¯=AB¯:DB¯=9,74,3210...DB¯=5,3789...

    Great: that gets us the value for side DB¯. Now we need to find the length of side CD¯. For that we will use the sine ratio (you can also do this calculation with the theorem of Pythagoras, but here we will do it with trigonometry).

    sinθ=oppositehypotenusesin(58,2°)=CD¯8,2(8,2)sin58,2°=CD¯(8,2)(0,8498...)=CD¯6,9691...=CD¯

    STEP: Calculate the final answer
    [−2 points ⇒ 0 / 7 points left]

    Now we can finally work out the angle that we want. CD¯=6,9691... and DB¯=5,3789... are the opposite and adjacent sides for angle B^, respectively, so this is a tangent ratio situation.

    tanθ=oppositeadjacenttanB^=6,9691...5,3789...B^=tan1(6,9691...5,3789...)=52,3380...52,3°

    Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

    The final answer is: B^=52,3°.


    Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows a scalene triangle. Two sides and an angle are given: AC¯=6,2, AB¯=10,9 and A^=36,8°. Point D is on side AB¯ as labelled, such that CD¯ makes a right angle with AB¯.

  1. Which of the following statements must be true about the figure above?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Imagine walking from point A to D and then walking on from D to B. How does this compare to walking straight from A to B without stopping?


    STEP: Consider the choices and select the correct option
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to select the true statement from the choices in the list. The correct option is about the sum of the segments along side AB¯. Point D breaks AB¯ up into two pieces. Therefore we know that together they make the total length of AB¯. Note that we do not know the exact position of point D, only that it is somewhere between A and B.

    The correct choice from the list is: Side AB is the sum of sides AD and DB.


    Submit your answer as:
  2. Compute the measure of B^. Round your answer to one decimal place.

    Answer: The size of B^ is °.
    numeric
    HINT: <no title>
    [−0 points ⇒ 7 / 7 points left]

    Use the segment CD¯ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.


    STEP: Draw a line to create right-angled triangles
    [−1 point ⇒ 6 / 7 points left]

    The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

    The line segment DC¯ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:


    STEP: Use trigonometry to find useful information
    [−4 points ⇒ 2 / 7 points left]

    In ΔADC (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

    With the information given, we can find both segments AD¯ and CD¯. Remember that we want to get the measure of B^, and for that we need both of these lengths. Start by calculating the length of AD¯, which will allow us to find DB¯ (because we know that AB¯=10,9). This calculation involves the hypotenuse and the side adjacent to A^, so use the cosine ratio.

    cosθ=adjacenthypotenusecos(36,8°)=AD¯6,2(6,2)cos36,8°=AD¯(6,2)(0,8007...)=AD¯4,9645...=AD¯SinceAD¯+DB¯=AB¯:DB¯=10,94,9645...DB¯=5,9354...

    Great: that gets us the value for side DB¯. Now we need to find the length of side CD¯. For that we will use the sine ratio (you can also do this calculation with the theorem of Pythagoras, but here we will do it with trigonometry).

    sinθ=oppositehypotenusesin(36,8°)=CD¯6,2(6,2)sin36,8°=CD¯(6,2)(0,5990...)=CD¯3,7139...=CD¯

    STEP: Calculate the final answer
    [−2 points ⇒ 0 / 7 points left]

    Now we can finally work out the angle that we want. CD¯=3,7139... and DB¯=5,9354... are the opposite and adjacent sides for angle B^, respectively, so this is a tangent ratio situation.

    tanθ=oppositeadjacenttanB^=3,7139...5,9354...B^=tan1(3,7139...5,9354...)=32,0350...32°

    Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

    The final answer is: B^=32°.


    Submit your answer as:

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One angle is labelled 37° and one side is 10 units long. Another one of the sides is labelled x.

Calculate the value of x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite side and the hypotenuse. So the trigonometric ratio to use is sin. We can write an equation with the given information and x:

sinθ=oppositehypotenusesin37°=x10

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Multiply by 10 to isolate x. Then evaluate the left side of the equation.

sin37°=x10(10)sin37°=x(10)(0,60181...)=x6,01815...=x6,02x

Therefore x is 6,02.


Submit your answer as:

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One angle is labelled 61° and one side is 6 units long. Another one of the sides is labelled x.

Solve for x.

INSTRUCTION: Round your answer to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite side and the hypotenuse. So the trigonometric ratio to use is sin. We can write an equation with the given information and x:

sinθ=oppositehypotenusesin61°=x6

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Multiply by 6 to isolate x. Then evaluate the left side of the equation.

sin61°=x6(6)sin61°=x(6)(0,87461...)=x5,24771...=x5,25x

Therefore x is 5,25.


Submit your answer as:

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One angle is labelled 51° and one side is 7 units long. Another one of the sides is labelled x.

Determine the value of x.

INSTRUCTION: Give your answer rounded to two decimal places.
Answer: x=
one-of
type(numeric.abserror(0.005))
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.


STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite side and the hypotenuse. So the trigonometric ratio to use is sin. We can write an equation with the given information and x:

sinθ=oppositehypotenusesin51°=x7

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for x. Multiply by 7 to isolate x. Then evaluate the left side of the equation.

sin51°=x7(7)sin51°=x(7)(0,77714...)=x5,44002...=x5,44x

Therefore x is 5,44.


Submit your answer as:

2. Special angles

Proportionality in the trigonometric ratios

Right-angled triangle PRQ is shown in the diagram below. Side PR has length 4, and angle Q^ has a measure of 30°.

Without using a calculator, determine the length of side RQ.

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.
Answer: RQ= .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The key to answering this question is the value of the given angle - it is one of the special angles! Use the known values of the trigonometric ratios of the special angles to set up an equation to solve for RQ.


STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side PR is opposite the given angle. And the unknown side RQ is adjacent to the angle. So we can write the following trigonometric equation:

tan30°=4RQ

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

Let's take a closer look at the 30° angle. Is there anything special about it? Yes - it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

tan30°=13

Since the left hand side of both of these equations is the same trigonometric ratio of 30°, we can construct the following proportion:

4RQ=13
NOTE:

This equation is a proportion. It represents the fact that any triangle with the same angle values will have the same ratio values, no matter what the lengths of the sides are. Here is the original triangle, but with another triangle which has the same angles:

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is, it only cares about how big the angles are because that determines the ratio of the sides.


STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side RQ. The value we want is in the denominator, so we can flip both fractions upside down to get it into the numerator.

4RQ=13RQ4=3RQ=34RQ=43

So the final answer is RQ=43.


Submit your answer as:

Special angles and equations

Given the following trigonometric equation:

8tan(x20°)=83

What is the value of x, if 0°x20°90°.

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by simplifying and removing any coefficients. What remains on the right hand side? The trigonometric ratio on the left hand side means that you can think of the right hand side as the ratio of two sides of a triangle.


STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for x in the following equation:

8tan(x20°)=83

We need to rearrange and simplify the equation until we reach a point where we know the value of x.

The first step is to remove the coefficient on the left hand side of the equation:

8tan(x20°)=83188tan(x20°)=1883tan(x20°)=3

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

We now have an equation with a trigonometric expression on the left hand side. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. We can rewrite the 3 as 31. Then we have information about two of the sides. And we know how to draw a triangle with sides of length 3 and 1. It is one of the special triangles!

Since the equation contains the function tan, we need to find the angle opposite the side of length 3 and adjacent to the side of length 1. This is shown in the diagram below:

We can see from the triangle that the angle we need is the 60°. So now we know:

tan(x20°)=3=tan60°

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for x. Since both sides of the equation contain the same trigonometric ratio, we can do the following:

tan(x20°)=tan60°x20°=60°

We can rearrange to solve for x:

x20°=60°x=60°+20°=80°
TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is x=80°.


Submit your answer as:

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 306090 triangle:

And the 454590 triangle:

In the first triangle, the acute angles are equal to 30° and 60°. In the second triangle, both of the acute angles are equal to 45°.

So the correct answer is the special angles are 30°, 45°, and 60°.


Submit your answer as:

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.
30°45°60°
sin A 12 32
cos 32 12 12
tan 13 1 B
Answer:
  • A =
  • B =
expression
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.


STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The question asks us to fill in the missing trigonometric ratios of the special angles. The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below. The 306090 triangle is:

And the 454590 triangle is:

One of the values that we need to find is sin30°. Looking at the 306090 triangle, we can see that the side opposite the 30° angle has length 1 and the hypotenuse has length 2. Taking the ratio of these gives us

sin30°=12

We can do the same thing to find tan60°. Looking at the 306090 triangle, we can see that the side opposite the 60° angle has length 3 and the side adjacent to it has length 1. Taking the ratio of these gives us

tan60°=3

So the completed table looks like this:

30°45°60°
sin 12 12 32
cos 32 12 12
tan 13 1 3

Submit your answer as: and

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than 90°.

Answer: One of the special angles is °.
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

There are three different special angles. These are the acute angles in the special triangles. If you can't remember what the special triangles look like, you can see them in the Everything Maths textbook here.


STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are 30°, 45°, and 60°. You could have given any one of these values as your answer.

Any one of these three angles is acceptable: 30°, 45°, or 60°.


Submit your answer as:

Special angles: what is the angle?

Given the following trigonometric equation:

cosx=12

What is the value of x?

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The left hand side of the equation is a trigonometric expression. This means that you can think of the right hand side as the ratio of two sides of a triangle. What triangle do you know that has sides of length 1 and 2? The answer will be one of the angles of this triangle.


STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for x in the following equation:

cosx=12

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

The left hand side of the equation is a trigonometric ratio. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with two sides of length 1 and 2. It is one of the special triangles!


STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio cos, we need to find the angle adjacent to the side of length 1. This is shown in the diagram below.

We can see from the triangle that the angle we need is the 45° angle.

So the answer is x=45°.


Submit your answer as:

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 306090 triangle:

And the 454590 triangle:

In the first triangle, the acute angles are equal to 30° and 60°. In the second triangle, both of the acute angles are equal to 45°.

Only one of these special angles was in the list of possible answers: 45°.

So the value of one of the special angles is 45°.


Submit your answer as:

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

12

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.
Answer: 12=
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing the special triangle that has sides of length 1 and 2. Identify which sides are opposite and adjacent to each of the acute angles. Then find which trigonometric ratio you need to use to get the expression given in the question.


STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

12

When a question asks us to work with a special angle, it is always a good first step to draw the corresponding special triangle. We do not yet know which angle we need, but we do know something about the sides of the triangle.

We know that one of the sides has length 1 and another has length 2. This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to 30° and 60°.


STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Consider the angle 30°. The side with length 1 is opposite 30°, and the side with length 2 is the hypotenuse (it is always opposite the right-angle). This is shown in the diagram below:

Since sin is defined as the ratio of the opposite and the hypotenuse, we can see that

12=sin30°
NOTE: The side of length 1 is also adjacent to the 60° angle, so 12=cos60° is also a valid solution.

So the answer is 12=sin30°.


Submit your answer as:

Using the special angles to solve problems

Quadrilateral KLMN, constructed from two triangles, is shown in the diagram below. Two angles are labelled: MKL=30° and KNM=45°. Side KL has length 3.

Find the length of side KN, marked with a ? in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.
Answer: The length of KN=
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side KM. Since MKL=30° is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.


STEP: Find the length of side KM
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side KN in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle KLM, we have angle MKL and side KL. We first want to find the length of side KM. So we can use the cosine ratio to write this equation:

cos30°=3KM

Let's take a closer look at the 30° angle. Is there anything special about it? Yes: it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

cos30°=32

Since the left hand sides of both of these equations are identical, we can solve for the length of side KM:

3KM=32KM=323KM=23

Now that we have some more information about the quadrilateral, let's add it to the diagram:


STEP: Find the length of side KN
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle KNM. So we can use trigonometry to find the length of side KN.

We can write the following trigonometric expression for triangle KNM:

sin45°=23KN

But 45° is also one the special angles! So we can use the same method of proportions to find side KN that we used to find side KM. We first write another equation for sin45°:

sin45°=12

And then use this to solve for KN:

23KN=12KN=232KN=26

So the correct answer is side KN=26.


Submit your answer as:

Trigonometry: special angles

Evaluate the following trigonometric expression:

tan60°
INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

ABCDE
12112332
Answer: The correct answer is option .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle 60°, which is a special angle! This means that we can evaluate the ratio tan60° exactly. If you can't remember the value of tan60°, start by drawing the special triangle which has the angle 60° in it. Then use the lengths of the sides of the special triangle to evaluate the ratio tan60°.


STEP: Draw a special triangle and use it to evaluate the expression tan60°
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

tan60°

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, 60° is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle 60° is shown below. Since we want to find tan60°, we need to identify the opposite and the adjacent.

Taking the ratio of the opposite and the adjacent, we can see that

tan60°=3
NOTE: You need to know the trigonometric ratios of the special angles for tests and exams. If you can't remember all of the ratios, start by drawing the correct special triangle. Then evaluate the ratio using the picture.

The correct answer is D: tan60°=3.


Submit your answer as:

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are both equal to 45°. Side BC¯ has length 1 and side CA¯ has length 2.

Answer the following questions about this triangle.

  1. What is the length of the missing side AB¯?

    INSTRUCTION: Type sqrt( ) if you need to show a square root.
    Answer: AB¯=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.


    STEP: Remember the values of the sides of the special triangle
    [−1 point ⇒ 0 / 1 points left]

    The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

    Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

    The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

    So the length of the missing side is AB¯=1.


    Submit your answer as:
  2. Using the triangle from Question 1, evaluate the following trigonometric expression:

    cos45°
    INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.
    Answer: cos45°=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio cos45° directly from the triangle.


    STEP: Evaluate the trigonometric ratio from the triangle
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to evaluate the trigonometric expression

    cos45°

    One way to do this would be to type this into a calculator. But the instructions say no calculators!

    In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate cos45° directly from the triangle by taking the ratio of the adjacent and the hypotenuse for the 45° angle. These sides are shown in the diagram below.

    Taking the ratio of the adjacent and the hypotenuse, we can see that

    cos45°=12
    NOTE: It does not matter which angle we use, because they are both 45°. If we use the other 45° angle, the answer would be the same.

    So the correct answer is cos45°=12.


    Submit your answer as:

Exercises

Proportionality in the trigonometric ratios

Right-angled triangle PQR is shown in the diagram below. Side QR has length 5, and angle R^ has a measure of 45°.

Without using a calculator, determine the length of side RP.

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.
Answer: RP= .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The key to answering this question is the value of the given angle - it is one of the special angles! Use the known values of the trigonometric ratios of the special angles to set up an equation to solve for RP.


STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side QR is adjacent to the given angle. And the unknown side RP is the hypotenuse of the triangle. So we can write the following trigonometric equation:

cos45°=5RP

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

Let's take a closer look at the 45° angle. Is there anything special about it? Yes - it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

cos45°=12

Since the left hand side of both of these equations is the same trigonometric ratio of 45°, we can construct the following proportion:

5RP=12
NOTE:

This equation is a proportion. It represents the fact that any triangle with the same angle values will have the same ratio values, no matter what the lengths of the sides are. Here is the original triangle, but with another triangle which has the same angles:

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is, it only cares about how big the angles are because that determines the ratio of the sides.


STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side RP. The value we want is in the denominator, so we can flip both fractions upside down to get it into the numerator.

5RP=12RP5=2RP=25RP=52

So the final answer is RP=52.


Submit your answer as:

Proportionality in the trigonometric ratios

Right-angled triangle XYZ is shown in the diagram below. Side XY has length 5, and angle X^ has a measure of 45°.

Without using a calculator, determine the length of side ZX.

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.
Answer: ZX= .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The key to answering this question is the value of the given angle - it is one of the special angles! Use the known values of the trigonometric ratios of the special angles to set up an equation to solve for ZX.


STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side XY is adjacent to the given angle. And the unknown side ZX is the hypotenuse of the triangle. So we can write the following trigonometric equation:

cos45°=5ZX

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

Let's take a closer look at the 45° angle. Is there anything special about it? Yes - it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

cos45°=12

Since the left hand side of both of these equations is the same trigonometric ratio of 45°, we can construct the following proportion:

5ZX=12
NOTE:

This equation is a proportion. It represents the fact that any triangle with the same angle values will have the same ratio values, no matter what the lengths of the sides are. Here is the original triangle, but with another triangle which has the same angles:

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is, it only cares about how big the angles are because that determines the ratio of the sides.


STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side ZX. The value we want is in the denominator, so we can flip both fractions upside down to get it into the numerator.

5ZX=12ZX5=2ZX=25ZX=52

So the final answer is ZX=52.


Submit your answer as:

Proportionality in the trigonometric ratios

Right-angled triangle XYZ is shown in the diagram below. Side ZX has length 22, and angle X^ has a measure of 60°.

Without using a calculator, determine the length of side YZ.

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.
Answer: YZ= .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The key to answering this question is the value of the given angle - it is one of the special angles! Use the known values of the trigonometric ratios of the special angles to set up an equation to solve for YZ.


STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side ZX is the hypotenuse of the triangle. And the unknown side YZ is opposite the angle. So we can write the following trigonometric equation:

sin60°=YZ22

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

Let's take a closer look at the 60° angle. Is there anything special about it? Yes - it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

sin60°=32

Since the left hand side of both of these equations is the same trigonometric ratio of 60°, we can construct the following proportion:

YZ22=32
NOTE:

This equation is a proportion. It represents the fact that any triangle with the same angle values will have the same ratio values, no matter what the lengths of the sides are. Here is the original triangle, but with another triangle which has the same angles:

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is, it only cares about how big the angles are because that determines the ratio of the sides.


STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side YZ.

YZ22=32YZ=3222YZ=6

So the final answer is YZ=6.


Submit your answer as:

Special angles and equations

Given the following trigonometric equation:

8sin(x+15°)=4

What is the value of x, if 0°x+15°90°.

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by simplifying and removing any coefficients. What remains on the right hand side? The trigonometric ratio on the left hand side means that you can think of the right hand side as the ratio of two sides of a triangle.


STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for x in the following equation:

8sin(x+15°)=4

We need to rearrange and simplify the equation until we reach a point where we know the value of x.

The first step is to remove the coefficient on the left hand side of the equation:

8sin(x+15°)=4188sin(x+15°)=184sin(x+15°)=12

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

We now have an equation with a trigonometric expression on the left hand side. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with sides of length 1 and 2. It is one of the special triangles!

Since the equation contains the function sin, we need to find the angle opposite the side of length 1. This is shown in the diagram below:

We can see from the triangle that the angle we need is the 30°. So now we know:

sin(x+15°)=12=sin30°

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for x. Since both sides of the equation contain the same trigonometric ratio, we can do the following:

sin(x+15°)=sin30°x+15°=30°

We can rearrange to solve for x:

x+15°=30°x=30°15°=15°
TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is x=15°.


Submit your answer as:

Special angles and equations

Given the following trigonometric equation:

2cos(x35°)=22

What is the value of x, if 0°x35°90°.

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by simplifying and removing any coefficients. What remains on the right hand side? The trigonometric ratio on the left hand side means that you can think of the right hand side as the ratio of two sides of a triangle.


STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for x in the following equation:

2cos(x35°)=22

We need to rearrange and simplify the equation until we reach a point where we know the value of x.

The first step is to remove the coefficient on the left hand side of the equation:

2cos(x35°)=22122cos(x35°)=1222cos(x35°)=12

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

We now have an equation with a trigonometric expression on the left hand side. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with sides of length 1 and 2. It is one of the special triangles!

Since the equation contains the function cos, we need to find the angle adjacent to the side of length 1. This is shown in the diagram below:

We can see from the triangle that the angle we need is the 45°. So now we know:

cos(x35°)=12=cos45°

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for x. Since both sides of the equation contain the same trigonometric ratio, we can do the following:

cos(x35°)=cos45°x35°=45°

We can rearrange to solve for x:

x35°=45°x=45°+35°=80°
TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is x=80°.


Submit your answer as:

Special angles and equations

Given the following trigonometric equation:

3tan(x+40°)=33

What is the value of x, if 0°x+40°90°.

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by simplifying and removing any coefficients. What remains on the right hand side? The trigonometric ratio on the left hand side means that you can think of the right hand side as the ratio of two sides of a triangle.


STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for x in the following equation:

3tan(x+40°)=33

We need to rearrange and simplify the equation until we reach a point where we know the value of x.

The first step is to remove the coefficient on the left hand side of the equation:

3tan(x+40°)=33133tan(x+40°)=1333tan(x+40°)=3

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

We now have an equation with a trigonometric expression on the left hand side. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. We can rewrite the 3 as 31. Then we have information about two of the sides. And we know how to draw a triangle with sides of length 3 and 1. It is one of the special triangles!

Since the equation contains the function tan, we need to find the angle opposite the side of length 3 and adjacent to the side of length 1. This is shown in the diagram below:

We can see from the triangle that the angle we need is the 60°. So now we know:

tan(x+40°)=3=tan60°

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for x. Since both sides of the equation contain the same trigonometric ratio, we can do the following:

tan(x+40°)=tan60°x+40°=60°

We can rearrange to solve for x:

x+40°=60°x=60°40°=20°
TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is x=20°.


Submit your answer as:

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 454590 triangle:

And the 306090 triangle:

In the first triangle, both of the acute angles are equal to 45°. In the second triangle, the acute angles are equal to 30° and 60°.

So the correct answer is the special angles are 30°, 45°, and 60°.


Submit your answer as:

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 454590 triangle:

And the 306090 triangle:

In the first triangle, both of the acute angles are equal to 45°. In the second triangle, the acute angles are equal to 30° and 60°.

So the correct answer is the special angles are 30°, 45°, and 60°.


Submit your answer as:

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 306090 triangle:

And the 454590 triangle:

In the first triangle, the acute angles are equal to 30° and 60°. In the second triangle, both of the acute angles are equal to 45°.

So the correct answer is the special angles are 30°, 45°, and 60°.


Submit your answer as:

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.
30°45°60°
sin 12 12 32
cos 32 A 12
tan 13 1 B
Answer:
  • A =
  • B =
expression
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.


STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The question asks us to fill in the missing trigonometric ratios of the special angles. The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below. The 306090 triangle is:

And the 454590 triangle is:

One of the values that we need to find is cos45°. Looking at the 454590 triangle, we can see that the side adjacent to the 45° angle has length 1 and the hypotenuse has length 2. Taking the ratio of these gives us

cos45°=12

We can do the same thing to find tan60°. Looking at the 306090 triangle, we can see that the side opposite the 60° angle has length 3 and the side adjacent to it has length 1. Taking the ratio of these gives us

tan60°=3

So the completed table looks like this:

30°45°60°
sin 12 12 32
cos 32 12 12
tan 13 1 3

Submit your answer as: and

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.
30°45°60°
sin A 12 32
cos 32 B 12
tan 13 1 3
Answer:
  • A =
  • B =
expression
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.


STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The question asks us to fill in the missing trigonometric ratios of the special angles. The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below. The 306090 triangle is:

And the 454590 triangle is:

One of the values that we need to find is sin30°. Looking at the 306090 triangle, we can see that the side opposite the 30° angle has length 1 and the hypotenuse has length 2. Taking the ratio of these gives us

sin30°=12

We can do the same thing to find cos45°. Looking at the 454590 triangle, we can see that the side adjacent to the 45° angle has length 1 and the hypotenuse has length 2. Taking the ratio of these gives us

cos45°=12

So the completed table looks like this:

30°45°60°
sin 12 12 32
cos 32 12 12
tan 13 1 3

Submit your answer as: and

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.
30°45°60°
sin A 12 32
cos 32 12 12
tan 13 B 3
Answer:
  • A =
  • B =
expression
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.


STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The question asks us to fill in the missing trigonometric ratios of the special angles. The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below. The 306090 triangle is:

And the 454590 triangle is:

One of the values that we need to find is sin30°. Looking at the 306090 triangle, we can see that the side opposite the 30° angle has length 1 and the hypotenuse has length 2. Taking the ratio of these gives us

sin30°=12

We can do the same thing to find tan45°. Looking at the 454590 triangle, we can see that the side opposite the 45° angle has length 1 and the side adjacent to it has length 1. Taking the ratio of these gives us

tan45°=1

So the completed table looks like this:

30°45°60°
sin 12 12 32
cos 32 12 12
tan 13 1 3

Submit your answer as: and

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than 90°.

Answer: One of the special angles is °.
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

There are three different special angles. These are the acute angles in the special triangles. If you can't remember what the special triangles look like, you can see them in the Everything Maths textbook here.


STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are 30°, 45°, and 60°. You could have given any one of these values as your answer.

Any one of these three angles is acceptable: 30°, 45°, or 60°.


Submit your answer as:

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than 90°.

Answer: One of the special angles is °.
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

There are three different special angles. These are the acute angles in the special triangles. If you can't remember what the special triangles look like, you can see them in the Everything Maths textbook here.


STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are 30°, 45°, and 60°. You could have given any one of these values as your answer.

Any one of these three angles is acceptable: 30°, 45°, or 60°.


Submit your answer as:

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than 90°.

Answer: One of the special angles is °.
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

There are three different special angles. These are the acute angles in the special triangles. If you can't remember what the special triangles look like, you can see them in the Everything Maths textbook here.


STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are 30°, 45°, and 60°. You could have given any one of these values as your answer.

Any one of these three angles is acceptable: 30°, 45°, or 60°.


Submit your answer as:

Special angles: what is the angle?

Given the following trigonometric equation:

cosx=12

What is the value of x?

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The left hand side of the equation is a trigonometric expression. This means that you can think of the right hand side as the ratio of two sides of a triangle. What triangle do you know that has sides of length 1 and 2? The answer will be one of the angles of this triangle.


STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for x in the following equation:

cosx=12

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

The left hand side of the equation is a trigonometric ratio. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with two sides of length 1 and 2. It is one of the special triangles!


STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio cos, we need to find the angle adjacent to the side of length 1. This is shown in the diagram below.

We can see from the triangle that the angle we need is the 45° angle.

So the answer is x=45°.


Submit your answer as:

Special angles: what is the angle?

Given the following trigonometric equation:

sinx=12

What is the value of x?

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The left hand side of the equation is a trigonometric expression. This means that you can think of the right hand side as the ratio of two sides of a triangle. What triangle do you know that has sides of length 1 and 2? The answer will be one of the angles of this triangle.


STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for x in the following equation:

sinx=12

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

The left hand side of the equation is a trigonometric ratio. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with two sides of length 1 and 2. It is one of the special triangles!


STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio sin, we need to find the angle opposite the side of length 1. This is shown in the diagram below.

We can see from the triangle that the angle we need is the 45° angle.

So the answer is x=45°.


Submit your answer as:

Special angles: what is the angle?

Given the following trigonometric equation:

cosx=12

What is the value of x?

Answer: x= °
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The left hand side of the equation is a trigonometric expression. This means that you can think of the right hand side as the ratio of two sides of a triangle. What triangle do you know that has sides of length 1 and 2? The answer will be one of the angles of this triangle.


STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for x in the following equation:

cosx=12

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

The left hand side of the equation is a trigonometric ratio. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with two sides of length 1 and 2. It is one of the special triangles!


STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio cos, we need to find the angle adjacent to the side of length 1. This is shown in the diagram below.

We can see from the triangle that the angle we need is the 45° angle.

So the answer is x=45°.


Submit your answer as:

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 306090 triangle:

And the 454590 triangle:

In the first triangle, the acute angles are equal to 30° and 60°. In the second triangle, both of the acute angles are equal to 45°.

Only one of these special angles was in the list of possible answers: 30°.

So the value of one of the special angles is 30°.


Submit your answer as:

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 306090 triangle:

And the 454590 triangle:

In the first triangle, the acute angles are equal to 30° and 60°. In the second triangle, both of the acute angles are equal to 45°.

Only one of these special angles was in the list of possible answers: 45°.

So the value of one of the special angles is 45°.


Submit your answer as:

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.


STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The 454590 triangle:

And the 306090 triangle:

In the first triangle, both of the acute angles are equal to 45°. In the second triangle, the acute angles are equal to 30° and 60°.

Only one of these special angles was in the list of possible answers: 60°.

So the value of one of the special angles is 60°.


Submit your answer as:

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

12

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.
Answer: 12=
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing the special triangle that has sides of length 1 and 2. Identify which sides are opposite and adjacent to each of the acute angles. Then find which trigonometric ratio you need to use to get the expression given in the question.


STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

12

When a question asks us to work with a special angle, it is always a good first step to draw the corresponding special triangle. We do not yet know which angle we need, but we do know something about the sides of the triangle.

We know that one of the sides has length 1 and another has length 2. This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to 45°.


STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Consider the angle 45°. The side with length 1 is adjacent to 45°, and the side with length 2 is the hypotenuse (it is always opposite the right-angle). This is shown in the diagram below:

Since cos is defined as the ratio of the adjacent and the hypotenuse, we can see that

12=cos45°
NOTE: The side of length 1 is also opposite the 45° angle, so 12=sin45° is also a valid solution.

So the answer is 12=cos45°.


Submit your answer as:

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

12

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.
Answer: 12=
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing the special triangle that has sides of length 1 and 2. Identify which sides are opposite and adjacent to each of the acute angles. Then find which trigonometric ratio you need to use to get the expression given in the question.


STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

12

When a question asks us to work with a special angle, it is always a good first step to draw the corresponding special triangle. We do not yet know which angle we need, but we do know something about the sides of the triangle.

We know that one of the sides has length 1 and another has length 2. This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to 30° and 60°.


STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Consider the angle 30°. The side with length 1 is opposite 30°, and the side with length 2 is the hypotenuse (it is always opposite the right-angle). This is shown in the diagram below:

Since sin is defined as the ratio of the opposite and the hypotenuse, we can see that

12=sin30°
NOTE: The side of length 1 is also adjacent to the 60° angle, so 12=cos60° is also a valid solution.

So the answer is 12=sin30°.


Submit your answer as:

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

12

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.
Answer: 12=
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing the special triangle that has sides of length 1 and 2. Identify which sides are opposite and adjacent to each of the acute angles. Then find which trigonometric ratio you need to use to get the expression given in the question.


STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

12

When a question asks us to work with a special angle, it is always a good first step to draw the corresponding special triangle. We do not yet know which angle we need, but we do know something about the sides of the triangle.

We know that one of the sides has length 1 and another has length 2. This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to 30° and 60°.


STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Consider the angle 60°. The side with length 1 is adjacent to 60°, and the side with length 2 is the hypotenuse (it is always opposite the right-angle). This is shown in the diagram below:

Since cos is defined as the ratio of the adjacent and the hypotenuse, we can see that

12=cos60°
NOTE: The side of length 1 is also opposite the 30° angle, so 12=sin30° is also a valid solution.

So the answer is 12=cos60°.


Submit your answer as:

Using the special angles to solve problems

Quadrilateral KLMN, constructed from two triangles, is shown in the diagram below. Two angles are labelled: KML=30° and KNM=45°. Side LM has length 23.

Find the length of side KN, marked with a ? in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.
Answer: The length of KN=
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side KM. Since KML=30° is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.


STEP: Find the length of side KM
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side KN in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle KLM, we have angle KML and side LM. We first want to find the length of side KM. So we can use the cosine ratio to write this equation:

cos30°=23KM

Let's take a closer look at the 30° angle. Is there anything special about it? Yes: it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

cos30°=32

Since the left hand sides of both of these equations are identical, we can solve for the length of side KM:

23KM=32KM=2323KM=4

Now that we have some more information about the quadrilateral, let's add it to the diagram:


STEP: Find the length of side KN
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle KNM. So we can use trigonometry to find the length of side KN.

We can write the following trigonometric expression for triangle KNM:

sin45°=4KN

But 45° is also one the special angles! So we can use the same method of proportions to find side KN that we used to find side KM. We first write another equation for sin45°:

sin45°=12

And then use this to solve for KN:

4KN=12KN=42KN=42

So the correct answer is side KN=42.


Submit your answer as:

Using the special angles to solve problems

Quadrilateral ABCD, constructed from two triangles, is shown in the diagram below. Two angles are labelled: ADC=45° and CAB=30°. Side CD has length 6.

Find the length of side AB, marked with a ? in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.
Answer: The length of AB=
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side AC. Since ADC=45° is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.


STEP: Find the length of side AC
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side AB in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle ADC, we have angle ADC and side CD. We first want to find the length of side AC. So we can use the sine ratio to write this equation:

sin45°=AC6

Let's take a closer look at the 45° angle. Is there anything special about it? Yes: it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

sin45°=12

Since the left hand sides of both of these equations are identical, we can solve for the length of side AC:

AC6=12AC=612AC=3

Now that we have some more information about the quadrilateral, let's add it to the diagram:


STEP: Find the length of side AB
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle ABC. So we can use trigonometry to find the length of side AB.

We can write the following trigonometric expression for triangle ABC:

cos30°=AB3

But 30° is also one the special angles! So we can use the same method of proportions to find side AB that we used to find side AC. We first write another equation for cos30°:

cos30°=32

And then use this to solve for AB:

AB3=32AB=332AB=32

So the correct answer is side AB=32.


Submit your answer as:

Using the special angles to solve problems

Quadrilateral WXYZ, constructed from two triangles, is shown in the diagram below. Two angles are labelled: YWX=30° and WZY=45°. Side WX has length 23.

Find the length of side WZ, marked with a ? in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.
Answer: The length of WZ=
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side WY. Since YWX=30° is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.


STEP: Find the length of side WY
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side WZ in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle WXY, we have angle YWX and side WX. We first want to find the length of side WY. So we can use the cosine ratio to write this equation:

cos30°=23WY

Let's take a closer look at the 30° angle. Is there anything special about it? Yes: it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

cos30°=32

Since the left hand sides of both of these equations are identical, we can solve for the length of side WY:

23WY=32WY=2323WY=4

Now that we have some more information about the quadrilateral, let's add it to the diagram:


STEP: Find the length of side WZ
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle WZY. So we can use trigonometry to find the length of side WZ.

We can write the following trigonometric expression for triangle WZY:

sin45°=4WZ

But 45° is also one the special angles! So we can use the same method of proportions to find side WZ that we used to find side WY. We first write another equation for sin45°:

sin45°=12

And then use this to solve for WZ:

4WZ=12WZ=42WZ=42

So the correct answer is side WZ=42.


Submit your answer as:

Trigonometry: special angles

Evaluate the following trigonometric expression:

tan30°
INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

ABCDE
121321213
Answer: The correct answer is option .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle 30°, which is a special angle! This means that we can evaluate the ratio tan30° exactly. If you can't remember the value of tan30°, start by drawing the special triangle which has the angle 30° in it. Then use the lengths of the sides of the special triangle to evaluate the ratio tan30°.


STEP: Draw a special triangle and use it to evaluate the expression tan30°
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

tan30°

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, 30° is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle 30° is shown below. Since we want to find tan30°, we need to identify the opposite and the adjacent.

Taking the ratio of the opposite and the adjacent, we can see that

tan30°=13
NOTE: You need to know the trigonometric ratios of the special angles for tests and exams. If you can't remember all of the ratios, start by drawing the correct special triangle. Then evaluate the ratio using the picture.

The correct answer is E: tan30°=13.


Submit your answer as:

Trigonometry: special angles

Evaluate the following trigonometric expression:

cos60°
INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

ABCDE
12323112
Answer: The correct answer is option .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle 60°, which is a special angle! This means that we can evaluate the ratio cos60° exactly. If you can't remember the value of cos60°, start by drawing the special triangle which has the angle 60° in it. Then use the lengths of the sides of the special triangle to evaluate the ratio cos60°.


STEP: Draw a special triangle and use it to evaluate the expression cos60°
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

cos60°

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, 60° is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle 60° is shown below. Since we want to find cos60°, we need to identify the adjacent and the hypotenuse.

Taking the ratio of the adjacent and the hypotenuse, we can see that

cos60°=12
NOTE: You need to know the trigonometric ratios of the special angles for tests and exams. If you can't remember all of the ratios, start by drawing the correct special triangle. Then evaluate the ratio using the picture.

The correct answer is A: cos60°=12.


Submit your answer as:

Trigonometry: special angles

Evaluate the following trigonometric expression:

cos45°
INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

ABCDE
121321312
Answer: The correct answer is option .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle 45°, which is a special angle! This means that we can evaluate the ratio cos45° exactly. If you can't remember the value of cos45°, start by drawing the special triangle which has the angle 45° in it. Then use the lengths of the sides of the special triangle to evaluate the ratio cos45°.


STEP: Draw a special triangle and use it to evaluate the expression cos45°
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

cos45°

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, 45° is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle 45° is shown below. Since we want to find cos45°, we need to identify the adjacent and the hypotenuse.

Taking the ratio of the adjacent and the hypotenuse, we can see that

cos45°=12
NOTE: You need to know the trigonometric ratios of the special angles for tests and exams. If you can't remember all of the ratios, start by drawing the correct special triangle. Then evaluate the ratio using the picture.

The correct answer is E: cos45°=12.


Submit your answer as:

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are equal to 30° and 60°. Side ST¯ has length 1 and side TV¯ has length 3.

Answer the following questions about this triangle.

  1. What is the length of the missing side VS¯?

    INSTRUCTION: Type sqrt( ) if you need to show a square root.
    Answer: VS¯=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.


    STEP: Remember the values of the sides of the special triangle
    [−1 point ⇒ 0 / 1 points left]

    The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

    Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

    The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

    So the length of the missing side is VS¯=2.


    Submit your answer as:
  2. Using the triangle from Question 1, evaluate the following trigonometric expression:

    cos30°
    INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.
    Answer: cos30°=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio cos30° directly from the triangle.


    STEP: Evaluate the trigonometric ratio from the triangle
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to evaluate the trigonometric expression

    cos30°

    One way to do this would be to type this into a calculator. But the instructions say no calculators!

    In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate cos30° directly from the triangle by taking the ratio of the adjacent and the hypotenuse for the 30° angle. These sides are shown in the diagram below.

    Taking the ratio of the adjacent and the hypotenuse, we can see that

    cos30°=32

    So the correct answer is cos30°=32.


    Submit your answer as:

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are both equal to 45°. Side ST¯ has length 1 and side TV¯ has length 1.

Answer the following questions about this triangle.

  1. What is the length of the missing side VS¯?

    INSTRUCTION: Type sqrt( ) if you need to show a square root.
    Answer: VS¯=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.


    STEP: Remember the values of the sides of the special triangle
    [−1 point ⇒ 0 / 1 points left]

    The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

    Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

    The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

    So the length of the missing side is VS¯=2.


    Submit your answer as:
  2. Using the triangle from Question 1, evaluate the following trigonometric expression:

    cos45°
    INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.
    Answer: cos45°=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio cos45° directly from the triangle.


    STEP: Evaluate the trigonometric ratio from the triangle
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to evaluate the trigonometric expression

    cos45°

    One way to do this would be to type this into a calculator. But the instructions say no calculators!

    In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate cos45° directly from the triangle by taking the ratio of the adjacent and the hypotenuse for the 45° angle. These sides are shown in the diagram below.

    Taking the ratio of the adjacent and the hypotenuse, we can see that

    cos45°=12
    NOTE: It does not matter which angle we use, because they are both 45°. If we use the other 45° angle, the answer would be the same.

    So the correct answer is cos45°=12.


    Submit your answer as:

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are both equal to 45°. Side AB¯ has length 1 and side CA¯ has length 2.

Answer the following questions about this triangle.

  1. What is the length of the missing side BC¯?

    INSTRUCTION: Type sqrt( ) if you need to show a square root.
    Answer: BC¯=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.


    STEP: Remember the values of the sides of the special triangle
    [−1 point ⇒ 0 / 1 points left]

    The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

    Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

    The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

    So the length of the missing side is BC¯=1.


    Submit your answer as:
  2. Using the triangle from Question 1, evaluate the following trigonometric expression:

    tan45°
    INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.
    Answer: tan45°=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio tan45° directly from the triangle.


    STEP: Evaluate the trigonometric ratio from the triangle
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to evaluate the trigonometric expression

    tan45°

    One way to do this would be to type this into a calculator. But the instructions say no calculators!

    In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate tan45° directly from the triangle by taking the ratio of the opposite and the adjacent for the 45° angle. These sides are shown in the diagram below.

    Taking the ratio of the opposite and the adjacent, we can see that

    tan45°=1
    NOTE: It does not matter which angle we use, because they are both 45°. If we use the other 45° angle, the answer would be the same.

    So the correct answer is tan45°=1.


    Submit your answer as:

3. Practical applications

Exercises

4. Trigonometry on a unit circle

Trigonometry on the Cartesian plane

Suppose tan(θ)=512, and 0°θ180°. Answer the three questions below about this equation.

  1. The angle θ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to θ?

    Answer: The angle θ points into Quadrant .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Compare the information given to the CAST diagram.


    STEP: Compare the information given to the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to figure out which quadrant the angle θ points to. We can answer this question using the CAST diagram because we already know that tan(θ) is positive.

    tan(θ)=512

    The CAST diagram tells us that the tangent ratio is positive in both Quadrants I and III. We also know that θ is in a specific interval: 0°θ180°.

    The CAST diagram above shows that θ must point into Quadrant I. That is the only quadrant which has the correct sign (positive) and is also in the allowed interval.

    NOTE: If we did not know that 0°θ180° we could not know which quadrant the angle is in!

    The angle θ must point into Quadrant I.


    Submit your answer as:
  2. On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates (x;y) and the radius r. The equation tan(θ)=512 tells us about the values of y and x. For angle θ, what is the value of r?

    Answer: The value of r is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by drawing a diagram to represent the angle θ. Remember from Question 1 you know the angle must be in Quadrant I.


    STEP: Draw a diagram to represent the angle
    [−1 point ⇒ 2 / 3 points left]

    Start by drawing a triangle on the Cartesian plane to represent the angle θ. From Question 1, we know that θ must be in Quadrant I. We will put a point in that quadrant and label it P. We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the x-axis.

    NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant I. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

    STEP: Determine the values available from tan(θ)=512
    [−1 point ⇒ 1 / 3 points left]

    For angle θ, we know that the tangent ratio is equal to 512. Comparing this to the definition of the tangent ratio, we can read off the values of y and x:

    tan(θ)=512=yxyx=512This means:y=5x=12

    Now we can label two sides of the triangle, and both of the coordinates of Point P.

    NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point P can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

    STEP: Calculate the third side of the triangle
    [−1 point ⇒ 0 / 3 points left]

    Now we can calculate the value of r for angle θ. We do this using the reference triangle: use the theorem of Pythagoras.

    a2+b2=c2(12)2+(5)2=r2144+25=r2169=r2±13=r

    The hypotenuse is always positive, so we must throw out the negative choice.

    The value of r is 13.


    Submit your answer as:
  3. Use the result from Question 2 to determine the value of cos(θ).

    Answer: cos(θ)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to combine the result of Question 2 with the definition of cos(θ).


    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the cosine ratio on the Cartesian plane is:

    cos(θ)=xr

    We know both x and r for angle θ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to θ, which is Point P.

    cos(θ)=xr=1213
    TIP: Compare the sign of your answer to the CAST diagram. Point P is in Quadrant I. And the CAST diagram tells us that the cosine ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of cos(θ) is 1213.


    Submit your answer as:

The trigonometric ratios on the Cartesian plane

On the Cartesian plane the trigonometric ratios are defined in terms of the coordinates x and y, and the radius r. Complete the following equation to make it true (identify the value which belongs in place of the ?):

cosθ=?r
Answer: The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of cosine on the Cartesian plane.


STEP: Use the definition of cosine on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

We use the opposite, adjacent, and hypotenuse sides of a right-angled triangle to define the trigonometric ratios. But we can translate these definitions onto the Cartesian plane using a right-angled triangle on the Cartesian plane.

The diagram below shows a point on the circle at (x;y). The radius of the circle is r. The point makes an angle θ with the positive x-axis. The sides of the triangle have lengths x, y, and r.

Here are three key relationships in the triangle above:

  • The hypotenuse of the triangle is the radius of the circle.
  • The side of the triangle opposite to θ has a length y.
  • The side of the triangle adjacent to θ has a length x.

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

sinθ=oppositehypotenuseyrcosθ=adjacenthypotenusexrtanθ=oppositeadjacentyx

By comparing these to the equation in the question we can find the missing value.

The correct definition is cosθ=xr so the missing quantity is x.


Submit your answer as:

Reading ratios from the Cartesian plane

The diagram below shows Point P at (3;4). A triangle is also shown. The lengths of the triangle's sides are 3, 4, and 5, as labelled. This triangle contains an angle which is labelled θ. The angle α is also labelled between the positive x-axis around to the hypotenuse of the triangle. Answer the two questions which follow.

  1. Based on the diagram, what is the value of sinθ?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: sinθ=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start with the fact that the sine ratio is always the opposite over the hypotenuse. You can find those values on the triangle in the figure.


    STEP: Read the answer from the triangle in the figure
    [−1 point ⇒ 0 / 1 points left]

    We need to find the value of sinθ. The sine ratio is always the opposite over the hypotenuse. So we can find the answer from the labels on the triangle.

    Now write down the ratio:

    sinθ=45
    NOTE: For the angle θ, we needed only the labels in the triangle. This will be different in Question 2.

    The value of sinθ is 45.


    Submit your answer as:
  2. What is the value of sinα?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: sinα= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the sine ratio on the Cartesian plane.


    STEP: Write down the definition of the sine ratio on the Cartesian plane
    [−1 point ⇒ 1 / 2 points left]

    In Question 1 we were working with the angle θ. But now we are working with the angle α. So we need to use the definition of sine on the Cartesian plane. This is for two reasons:

    • Unlike θ, the angle α is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
    • The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive x-axis, and that is where the angle α starts.

    The definition for the sine ratio is:

    sinα=yr

    STEP: Substitute the values for y and r
    [−1 point ⇒ 0 / 2 points left]

    To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the y-coordinate of Point P, which is 4.

    sinα=yrFor Point P,y=4,r=5sinα=45

    It is now clear that sinα is not the same as sinθ. But the numbers in each ratio are the same: only the signs are different. This is because the angles are closely related: they are both connected to Point P. θ is called a reference angle for α. And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle α.

    NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for sinα depends on which quadrant the reference triangle is in (the size of α). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the sine ratio is negative in the fourth quadrant, which tells us that sinα must be negative.

    The value of sinα is 45.


    Submit your answer as:

Simplifying trigonometric expressions with reduction formulas

Without the use of a calculator, evaluate the trigonometric function shown.

sin(390°)
INSTRUCTIONS:
  • Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
  • Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.
Answer: sin(390°)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of 90°,180°, or 360°. The other part should be one of the special angles, 30°,45°, or 60°.


STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

  • We must show the working step by step.
  • We can expect to use special angles to solve the problem. The special angles are 30°,45°, and 60°.

The first step is to break the angle in the question into two terms so that we can use one of the reduction formulas. We want one of those pieces to be a special angle because that is the only way we can evaluate the expression without a calculator. In this case, the original angle is 390°. We can split that into the values 360° and 30° using addition.

sin(390°)=sin(360°+30°)

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. Specifically, we can remove the 360° part of the angle using the formula:

sin(360°+θ)=sinθ
TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression sin(360°+30°). Note that there is a negative at the front of the expression. But it is not part of the reduction formula, so it will just sit there until the end.

sin(360°+30°)=sin(30°)

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

=sin(30°)=12

The final answer is sin(390°)=12.


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Special angles: 0° and 90°

  1. On the Cartesian plane, the cosine ratio is defined as xr. Based on this definition or otherwise, determine the value of cos0°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of cos0° is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of cos0°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate xr
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of cos0° is 1.


    Submit your answer as:
  2. Question 1 was about the value of cos0°. Which of the reasons below explains the answer to Question 1?

    Answer:

    The explanation for the answer to Question 1 is that when θ=0°, .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=0°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why cos0° is 1. The answer comes directly from the definition for the cosine ratio on the Cartesian plane. That definition is:

    cosθ=xr

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 0°. When that happens, the hypotenuse will rotate down. And the triangle gets more flat until the hypotenuse falls onto the horizontal side of the triangle on the x-axis.

    For θ=0°, the triangle collapses to a horizontal line segment on the x-axis. The segment reaches from the origin over to the circle at the point (x;0). And the following things are both true about the "triangle" represented by that line:

    • x=r
    • y=0

    Let's use this information in the definition of the cosine ratio on the Cartesian plane:

    cosθ=xrcos0°=rr=1

    And finally we can see why cos0° must be 1: it is because when the angle is zero, the side of the triangle corresponding to x is the same length as the radius.

    The reason cos0° is 1 is because x=r.


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Special angles: 180°, 270°, and 360°

  1. On the Cartesian plane, the tangent function is defined as yx. Based on this definition or otherwise, determine the value of tan270°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of tan270° is .
    string
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of tan270°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate yx
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of tan270° is undefined.


    Submit your answer as:
  2. Question 1 was about the value of tan270°. Which of the reasons below explains the answer to Question 1?

    Answer: The explanation for the answer to Question 1 is that when θ=270°, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=270°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why tan270° is undefined. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

    tanθ=yx

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 270°. When that happens, the hypotenuse will rotate into Quadrant II and then into Quadrant III, coming to a rest on the negative y-axis (pointing straight down). This is shown below.

    For θ=270°, the triangle collapses to a vertical line segment on the y-axis. The segment reaches from the origin down to the circle at the point (0;y). And the following things are both true about the "triangle" represented by that line:

    • x=0
    • y=r

    Remember that r, the radius, is positive. But for θ=270° we know that y must be negative (you can see this on the diagram above). That is the reason why y=r, not y=r.

    Now we can use this information in the definition of the tangent ratio on the Cartesian plane:

    tanθ=yxtan270°=y0=undefined

    And finally we can see why tan270° must be undefined: it is because when the angle is 270°, the side of the triangle corresponding to x shrinks to zero, leading to division by zero.

    The reason tan270° is undefined is because x=0.


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Trigonometry gone wild!

  1. Point M is given on the Cartesian plane with the origin at O. It is in the second quadrant at (16;yM), where yM is the y-coordinate of Point M. M makes an angle δ with the positive x-axis, as labelled, and OM¯ is 34 units long. Given that sinδ=1517, determine the value of yM. The diagram may or may not be drawn to scale.

    Answer: yM =
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the sine ratio on the Cartesian plane to write an equation which includes the y-coordinate.


    STEP: Write an equation based on the information given
    [−1 point ⇒ 1 / 2 points left]

    We need to determine the y-coordinate of Point M. The point sits at an angle of δ to the positive x-axis and we know that sinδ is equal to 1517. As always, the sine ratio is defined as yr. So we can write:

    yr=1517

    This equation is true for the angle δ and Point M. So we can substitute in the radius value of 34 on the left-hand side of the equation. (We will also change y to yM because the equation is specifically about Point M now.)

    yM(34)=1517
    NOTE:

    The equation above is a proportion. It represents the fact that any point at an angle of δ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle δ. The second point makes a smaller triangle.

    The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!


    STEP: Solve the equation
    [−1 point ⇒ 0 / 2 points left]

    Now we need to solve the equation. Multiply both sides by the denominator on the left side to isolate yM.

    yM(34)=1517yM=1517(34)yM=30
    TIP: Check the sign of the answer: it must be positive because Point M is in the second quadrant. If your sign is wrong, you should go back and check through your work.

    The y-coordinate of Point M is 30.


    Submit your answer as:
  2. Evaluate the following expression:

    518cos2δ+1
    INSTRUCTION: Your answer must be a simplified fraction.
    Answer: 518cos2δ+1=
    fraction
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the result of Question 1 to calculate the value of cosδ. Then evaluate the expression.


    STEP: Determine the value of cosδ
    [−1 point ⇒ 1 / 2 points left]

    We can start by finding the value of cosδ. From Question 1 we know everything about Point M (which is linked to the angle δ):

    On the Cartesian plane the cosine ratio is xr. Using the values from the figure above we get:

    cosθ=xrcosδ=(16)(34)=817

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now we can evaluate the expression. Substitute 817 in for cosδ and simplify:

    518cos2δ+1=518(817)2+1=518(64289)+1=2417+1=2417+1717=4117

    The value of the expression is 4117.


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Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter T belong?

Answer:

The T should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

The letters sit on the Cartesian plane as shown below: C is in the lower right quadrant (Quadrant IV). The letters C-A-S-T go around the diagram in the anticlockwise direction. You will find it helpful to memorise the positions of the letters!

The T belongs in Quadrant III. The T in Quadrant III tells us that the tangent ratio is positive for any angle in Quadrant III. At the same time, it tells us that the sine and cosine ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The T should be in Quadrant III.


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Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle θ:

sinθ=513cosθ=1213

We can draw a reference triangle on the Cartesian plane to represent the angle θ. One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).
Answer: The coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the sine ratio is positive and the second equation shows that the cosine ratio is positive. Use these two facts and the CAST diagram to figure out which quadrant the angle θ must point into.


STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We need to determine the coordinates of the point on the Cartesian plane which correspond to the reference triangle for the angle θ. The first thing we need to do is figure out which quadrant the angle points into. Then we can draw a point and the reference triangle we need in that quadrant.

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle θ must point into. The first equation shows that the sine ratio is positive and the second equation shows that the cosine ratio is positive. So the point must be in Quadrant I: that is the only quadrant where the sine and the cosine ratios are both positive.


STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant I, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant I.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of x and y from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the sine and cosine ratios.

The definition for sine is yr. The y in this definition is one of the coordinates we want!

sinθ=513yrwhich means:y=5r=13

The definition for cosine is xr. And we can follow the same logic as above to find the value of x.

cosθ=1213xrwhich means:x=12r=13

Now we know that the coordinates of the point in the reference triangle are (12;5). And we also know the radius, r, for the reference triangle. The reference triangle looks like this:

NOTE: While the lengths of the triangle's sides must be positive, the coordinates of the point can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

The coordinates of the point in the reference triangle are (12;5).


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Trigonometric ratios on the Cartesian plane: positives & negatives

  1. Consider the following equation about an angle θ:

    cosθ=1213

    The diagram below shows four points on the Cartesian plane. They are labelled W,X,Y, and Z. The diagram is not drawn to scale.

    Two of the four points above correspond to the equation cosθ=1213. Which two?

    Answer:

    The points corresponding to cosθ=1213 are Points .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to compare the sign of the ratio to the CAST diagram.


    STEP: Use the CAST diagram and the sign of the ratio given
    [−1 point ⇒ 0 / 1 points left]

    On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

    In this case we know that:

    cosθ=1213negative

    This equation tells us that the angle θ must point into one of the two quadrants where the cosine ratio is negative. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

    We are dealing with the cosine ratio, and we know that the ratio has a negative value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. So we do not want Point W. Similarly, the C in Quadrant IV means that the cosine ratio is positive in that quadrant so we do not want Point Z either. So Points X and Y are the points which correspond to the equation in the question.

    The correct points are X and Y.


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  2. In Question 1 you found the two points which correspond to this equation:

    cosθ=1213

    There are two values for θ connected to those points. From the list of choices below, select the two values closest to θ (the choices in the list have been rounded).

    TIP: Use the answer to Question 1.
    Answer: The values for θ are closest to: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to pick the angles which match the correct points in Question 1.


    STEP: Use the answer to Question 1
    [−1 point ⇒ 0 / 1 points left]

    For this question we need to identify the values of θ which correspond to the equation

    cosθ=1213

    There are two such values: they match the points we identified in Question 1.

    We already know that Points X and Y correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: 23°;157°;203°;337°. Two of these choices agree with the positions of Points X and Y. (Remember that the angles always start on the positive x-axis, as shown below.)

    We do not know the exact values of these two angles. But we know that one of them points into Quadrant II and the other points into Quadrant III. That means the smaller angle must be obtuse (between 90° and 180°) and the larger angle must be between 180° and 270°. Based on the choices in the list, the values for θ must be 157° and 203°.

    The two values for θ are 157° and 203°.


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Finding a ratio from a point

The figure below shows Point A at (5;y). It is a distance 13 from the origin, as labelled. The angle between the positive x-axis and Point A is B. This diagram is not drawn to scale.

Answer the two question which follow about Point A and angle B.

  1. Compute the y-coordinate of Point A.

    Answer: The y-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.


    STEP: Draw a right-angled triangle onto the given figure
    [−1 point ⇒ 2 / 3 points left]

    We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the x-coordinate of Point A.

    Notice that the side of the triangle is labelled with a positive number, even though the x-coordinate is negative. See the note below for an explanation about why these signs are not always the same.


    STEP: Use the triangle to calculate the missing side
    [−1 point ⇒ 1 / 3 points left]

    Remember that we want to find the y-coordinate of Point A. We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

    a2+b2=c2y2+(5)2=(13)2y2+25=169y2=144y=±144y=±12

    This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is 12 units.


    STEP: Check the sign and adjust it if necessary
    [−1 point ⇒ 0 / 3 points left]

    The result above tells us that Point A is 12 units away from the x-axis. So the missing coordinate must be either 12 or 12. It depends on the position of the point! From the diagram we can see that Point A is in Quadrant III, so the y-coordinate has to be negative.

    NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: 3 spaces right is not the same as 3 spaces left. It is crucial to be aware that these signs can be different.

    The y-coordinate of Point A is y=12.


    Submit your answer as:
  2. Now determine the value of sin(B).

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer: sin(B)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the coordinates of the point and the radius, whichever are necessary.


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the sine ratio on the Cartesian plane is:

    sin(θ)=yr

    Point A corresponds to angle B, so all we need to do is substitute in the correct values from the point. In this case that means using the y-coordinate and the radius.

    sin(θ)=yrsin(B)=1213
    TIP: Compare the sign of your answer to the CAST diagram. Point A is in Quadrant III. And the CAST diagram tells us that the sine ratio in Quadrant III is always negative. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(B) is 1213.


    Submit your answer as:

Finding one ratio from another

Given:

cos(C)=1314

Determine the value of tan(C) if C is greater than 180° and less than 360°.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).
Answer: tan(C)= .
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle C is in and draw a sketch to represent that angle.


STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle C points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is greater than 180° and less than 360°, and the fact that cos(C) is negative, the angle must be in Quadrant III. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the cosine ratio is the ratio of x to r. While the value of cos(C) is negative, the labels on the triangle are positive because lengths are always positive.

NOTE: The figure above is a sketch: the triangle is probably not actually the size shown above. That is not a problem, but it means that we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Find the value of tan(C)
[−2 points ⇒ 0 / 4 points left]

We want to find tan(C), which means we want the ratio yx. So we need the value of y. We can find this using the theorem of Pythagoras with the reference triangle:

a2+b2=c2y2+(13)2=(14)2y2=196169y2=27y2=93y2=±93y=±33

This calculation tells us two things: the third side of the triangle is 33 units long, and the y-coordinate of the point in our diagram is 33. Now we have enough information to evaluate the tangent ratio for angle C.

tan(C)=yx=3313=3313
TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant III and we expect the tangent ratio to be positive in this quadrant. So all is groovy!

The value of tan(C) is 3313.


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The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant I, where we find the letter A:

The A is always in Quadrant I. All three of the trigonometric ratios are positive for any angle in Quadrant I (any angle between 0° and 90°).

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the tangent ratio is positive in Quadrants I and III, while it is negative in Quadrants II and IV.

The CAST diagram tells us when the trigonometric ratios are positive or negative.


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Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point G is on the circle at (15;8). The angle between the positive x-axis and Point G is g. A radius of the circle is shown from the origin to Point G.

Answer the two questions which follow about Point G and g.

  1. Determine the length of the radius, r.

    Answer: The length of the radius is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by drawing a right-angled reference triangle on the figure. This triangle should include Point G and the radius shown.


    STEP: Draw a right-angled triangle on the figure
    [−1 point ⇒ 1 / 2 points left]

    We need to find the radius of the circle. We can do this by using the coordinates of Point G. This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point G, the origin, and the x-axis, as shown below.

    Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point G. The x-coordinate of Point G is 15: that means G is 15 units away from the y-axis. Similarly, the y-coordinate of G is 8, which means that G is 8 units away from the x-axis.

    TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

    STEP: Use the triangle to calculate the radius
    [−1 point ⇒ 0 / 2 points left]

    The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

    a2+b2=c2(15)2+(8)2=r2289=r2±289=r2±17=r

    We get a radius of ±17. But it must be positive because it is a length.

    The length of the radius of the circle is 17.


    Submit your answer as:
  2. Hence determine the value of sin(g).

    Answer: The value of sin(g) is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the definition of the sine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    We need to find the value of sin(g). The trigonometric functions are defined on the Cartesian plane in terms of the coordinates (x;y) and the radius r. And for g, we know all three of those numbers from Question 1! In this case we want to know sin(g), so we need the definition of the sine ratio on the Cartesian plane:

    sin(θ)=yr

    Point G corresponds to g, so we need to substitute in the correct values from the point. In this case that means using the y-coordinate and the radius.

    y=8r=17sin(g)=817
    TIP: Compare the sign of your answer to the CAST diagram. Point G is in Quadrant I. The CAST diagram tells us that the sine ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(g) is 817.


    Submit your answer as:

Finding one ratio from another

Suppose cos(ω)=725, and 180°ω360°. Without using a calculator, determine the value of tan(ω).

INSTRUCTION: Write your answer as a simplified fraction.
Answer: The value of tan(ω) is .
fraction
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which ω sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find tan(ω).


STEP: Determine which quadrant ω must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions cos(ω) and tan(ω). We know one and need to find the other. We can do this because cos(ω) tells us about two of the three values x, y, and r. We can use those to find the third value, which we need in order to find tan(ω).

The first step is to figure out which quadrant the angle ω points to. We can do this by combining the CAST diagram with the interval given in the question: 180°ω360°. We already know that cos(ω) is negative because it is equal to 725. The CAST diagram tells us that the cosine ratio is negative in Quadrants II and III. But since ω is between 180° and 360°, it cannot be in Quadrant II.

NOTE: If we did not know that 180°ω360°, we could not know if the angle is in Quadrant II or III.

The angle ω must point to Quadrant III.


STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant III to represent angle ω.

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant III so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that cos(ω) is equal to 725 (this is given in the question). This tells us about two sides of the triangle in the picture above.

cos(θ)=725=xrxr=725This means:x=7r=25

Now we know one of the coordinates of Point P. We also know two sides of the triangle. Note that the lengths of the triangle's sides must be positive because they are distances, no matter which quadrant the triangle is in.


STEP: Find the value of y
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of y. We can find this using the theorem of Pythagoras for the reference triangle in the figure.

a2+b2=c2y2+(7)2=(25)2y2=62549y2=576y=±24

For the third side of the triangle, we must take the positive value (because it is a distance). However, the sign of y depends on which quadrant the point is in. In this case, Point P is in Quadrant III, where the y-values must be negative. So the correct value for the y-coordinate is 24 (we throw away the positive answer).


STEP: Use the definition of the tangent ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point P corresponds to angle ω, we can evaluate tan(ω) by substituting in the correct values from the point.

tan(θ)=yx=247
TIP: Point P is in Quadrant III, where the CAST diagram tells us that the tangent ratio is always positive. Make sure your answer agrees with that!

The value of tan(ω) is 247.


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Working with trigonometric ratios

The diagram below shows Point P at (3;7). Triangle OPQ is also shown, with three sides labelled as 3, 7, and 58. The angle α reaches from the positive x-axis to the line OP, as labelled. Answer the two questions which follow below.

  1. Determine the value of cosα.

    INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).
    Answer: cosα=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition for the cosine ratio on the Cartesian plane:

    cosθ=xr

    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    On the Cartesian plane the cosine ratio is defined as xr. That means that we can read the answer from the diagram in the question. We need the values for x and r for Point P.

    P is at (3;7). From the diagram we can see that the radius is r=58. So we get:

    x=3andr=58

    Putting those numbers into the definition for the cosine ratio, we get:

    cosα=xrcosα=358
    NOTE: The size of angle α puts Point P in Quadrant IV. The CAST diagram tells us that the cosine ratio is always positive in Quadrant IV. And indeed the answer we got agrees with that: it is positive because the x-coordinate is positive (and the radius is always positive).

    The value of cosα is 358.


    Submit your answer as:
  2. Hence evaluate the following expression:

    4cos2α32
    Answer: 4cos2α32=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.


    STEP: Substitute the value of cosα into the expression
    [−1 point ⇒ 1 / 2 points left]

    Start by substituting the answer from Question 1 into the expression:

    4cos2α32=4(358)232

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now evaluate the expression.

    4(358)232=495832=182932=(182)(329)292=5158

    The value of 4cos2α32 is 5158.


    Submit your answer as:

Using the CAST diagram

  1. The figure below shows the CAST diagram. Point P is shown, and angle θ is the angle from the positive x-axis to Point P.

    Is sinθ positive or negative?

    Answer: sinθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Point P is in Quadrant II. What does the CAST diagram tell you about the sign of the sine ratio in that quadrant?


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to determine if the expression sinθ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

    We can see that the angle θ puts Point P in Quadrant II. On the CAST diagram, what letter is in Quadrant II? It is S. The S means that the sine ratio is positive in that quadrant and the other ratios are negative.

    Based on the CAST diagram, the expression sinθ must be positive.


    Submit your answer as:
  2. Now determine if tanθ is positive or negative.

    Answer: tanθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the same approach as you used in Question 1.


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    For this question, we use exactly the same information that we used in Question 1. Point P and θ are in Quadrant II, where we find the letter S on the CAST diagram. This means that the sine ratio is positive for any angle in that quadrant, while cosine and tangent are negative.

    NOTE:

    What do the answers to these questions mean?

    The angle shown in the diagram is θ=132°. We can calculate the value of the expressions for these two questions (use a calculator):

    From Question 1:sin132°=0,74314...From Question 2:tan132°=1,11061...

    You can see that these values are positive and negative, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

    The expression tanθ must be negative.


    Submit your answer as:

Exercises

Trigonometry on the Cartesian plane

Suppose cos(θ)=45, and 0°θ180°. Answer the three questions below about this equation.

  1. The angle θ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to θ?

    Answer: The angle θ points into Quadrant .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Compare the information given to the CAST diagram.


    STEP: Compare the information given to the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to figure out which quadrant the angle θ points to. We can answer this question using the CAST diagram because we already know that cos(θ) is negative.

    cos(θ)=45

    The CAST diagram tells us that the cosine ratio is negative in both Quadrants II and III. We also know that θ is in a specific interval: 0°θ180°.

    The CAST diagram above shows that θ must point into Quadrant II. That is the only quadrant which has the correct sign (negative) and is also in the allowed interval.

    NOTE: If we did not know that 0°θ180° we could not know which quadrant the angle is in!

    The angle θ must point into Quadrant II.


    Submit your answer as:
  2. On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates (x;y) and the radius r. The equation cos(θ)=45 tells us about the values of x and r. For angle θ, what is the value of the y-coordinate?

    Answer: The value of the y-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by drawing a diagram to represent the angle θ. Remember from Question 1 you know the angle must be in Quadrant II.


    STEP: Draw a diagram to represent the angle
    [−1 point ⇒ 2 / 3 points left]

    Start by drawing a triangle on the Cartesian plane to represent the angle θ. From Question 1, we know that θ must be in Quadrant II. We will put a point in that quadrant and label it P. We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the x-axis.

    NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant II. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

    STEP: Determine the values available from cos(θ)=45
    [−1 point ⇒ 1 / 3 points left]

    For angle θ, we know that the cosine ratio is equal to 45. Comparing this to the definition of the cosine ratio, we can read off the values of x and r:

    cos(θ)=45=xrxr=45This means:x=4r=5

    Now we can label two sides of the triangle, and one of the coordinates of Point P.

    NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point P can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

    STEP: Calculate the third side of the triangle
    [−1 point ⇒ 0 / 3 points left]

    Now we can calculate the value of y for angle θ. We do this using the reference triangle: use the theorem of Pythagoras.

    a2+b2=c2y2+(4)2=(5)2y2=2516y2=9y=±3

    The length of a side of a triangle is always positive. And since Point P is in Quadrant II, the y-coordinate is going to be positive as well.

    The value of the y-coordinate is 3.


    Submit your answer as:
  3. Use the result from Question 2 to determine the value of sin(θ).

    Answer: sin(θ)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to combine the result of Question 2 with the definition of sin(θ).


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the sine ratio on the Cartesian plane is:

    sin(θ)=yr

    We know both y and r for angle θ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to θ, which is Point P.

    sin(θ)=yr=35
    TIP: Compare the sign of your answer to the CAST diagram. Point P is in Quadrant II. And the CAST diagram tells us that the sine ratio in Quadrant II is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(θ) is 35.


    Submit your answer as:

Trigonometry on the Cartesian plane

Suppose cos(θ)=35, and 0°θ180°. Answer the three questions below about this equation.

  1. The angle θ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to θ?

    Answer: The angle θ points into Quadrant .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Compare the information given to the CAST diagram.


    STEP: Compare the information given to the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to figure out which quadrant the angle θ points to. We can answer this question using the CAST diagram because we already know that cos(θ) is positive.

    cos(θ)=35

    The CAST diagram tells us that the cosine ratio is positive in both Quadrants I and IV. We also know that θ is in a specific interval: 0°θ180°.

    The CAST diagram above shows that θ must point into Quadrant I. That is the only quadrant which has the correct sign (positive) and is also in the allowed interval.

    NOTE: If we did not know that 0°θ180° we could not know which quadrant the angle is in!

    The angle θ must point into Quadrant I.


    Submit your answer as:
  2. On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates (x;y) and the radius r. The equation cos(θ)=35 tells us about the values of x and r. For angle θ, what is the value of the y-coordinate?

    Answer: The value of the y-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by drawing a diagram to represent the angle θ. Remember from Question 1 you know the angle must be in Quadrant I.


    STEP: Draw a diagram to represent the angle
    [−1 point ⇒ 2 / 3 points left]

    Start by drawing a triangle on the Cartesian plane to represent the angle θ. From Question 1, we know that θ must be in Quadrant I. We will put a point in that quadrant and label it P. We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the x-axis.

    NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant I. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

    STEP: Determine the values available from cos(θ)=35
    [−1 point ⇒ 1 / 3 points left]

    For angle θ, we know that the cosine ratio is equal to 35. Comparing this to the definition of the cosine ratio, we can read off the values of x and r:

    cos(θ)=35=xrxr=35This means:x=3r=5

    Now we can label two sides of the triangle, and one of the coordinates of Point P.

    NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point P can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

    STEP: Calculate the third side of the triangle
    [−1 point ⇒ 0 / 3 points left]

    Now we can calculate the value of y for angle θ. We do this using the reference triangle: use the theorem of Pythagoras.

    a2+b2=c2y2+(3)2=(5)2y2=259y2=16y=±4

    The length of a side of a triangle is always positive. And since Point P is in Quadrant I, the y-coordinate is going to be positive as well.

    The value of the y-coordinate is 4.


    Submit your answer as:
  3. Use the result from Question 2 to determine the value of sin(θ).

    Answer: sin(θ)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to combine the result of Question 2 with the definition of sin(θ).


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the sine ratio on the Cartesian plane is:

    sin(θ)=yr

    We know both y and r for angle θ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to θ, which is Point P.

    sin(θ)=yr=45
    TIP: Compare the sign of your answer to the CAST diagram. Point P is in Quadrant I. And the CAST diagram tells us that the sine ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(θ) is 45.


    Submit your answer as:

Trigonometry on the Cartesian plane

Suppose cos(θ)=2425, and 0°θ180°. Answer the three questions below about this equation.

  1. The angle θ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to θ?

    Answer: The angle θ points into Quadrant .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Compare the information given to the CAST diagram.


    STEP: Compare the information given to the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to figure out which quadrant the angle θ points to. We can answer this question using the CAST diagram because we already know that cos(θ) is negative.

    cos(θ)=2425

    The CAST diagram tells us that the cosine ratio is negative in both Quadrants II and III. We also know that θ is in a specific interval: 0°θ180°.

    The CAST diagram above shows that θ must point into Quadrant II. That is the only quadrant which has the correct sign (negative) and is also in the allowed interval.

    NOTE: If we did not know that 0°θ180° we could not know which quadrant the angle is in!

    The angle θ must point into Quadrant II.


    Submit your answer as:
  2. On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates (x;y) and the radius r. The equation cos(θ)=2425 tells us about the values of x and r. For angle θ, what is the value of the y-coordinate?

    Answer: The value of the y-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by drawing a diagram to represent the angle θ. Remember from Question 1 you know the angle must be in Quadrant II.


    STEP: Draw a diagram to represent the angle
    [−1 point ⇒ 2 / 3 points left]

    Start by drawing a triangle on the Cartesian plane to represent the angle θ. From Question 1, we know that θ must be in Quadrant II. We will put a point in that quadrant and label it P. We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the x-axis.

    NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant II. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

    STEP: Determine the values available from cos(θ)=2425
    [−1 point ⇒ 1 / 3 points left]

    For angle θ, we know that the cosine ratio is equal to 2425. Comparing this to the definition of the cosine ratio, we can read off the values of x and r:

    cos(θ)=2425=xrxr=2425This means:x=24r=25

    Now we can label two sides of the triangle, and one of the coordinates of Point P.

    NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point P can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

    STEP: Calculate the third side of the triangle
    [−1 point ⇒ 0 / 3 points left]

    Now we can calculate the value of y for angle θ. We do this using the reference triangle: use the theorem of Pythagoras.

    a2+b2=c2y2+(24)2=(25)2y2=625576y2=49y=±7

    The length of a side of a triangle is always positive. And since Point P is in Quadrant II, the y-coordinate is going to be positive as well.

    The value of the y-coordinate is 7.


    Submit your answer as:
  3. Use the result from Question 2 to determine the value of sin(θ).

    Answer: sin(θ)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to combine the result of Question 2 with the definition of sin(θ).


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the sine ratio on the Cartesian plane is:

    sin(θ)=yr

    We know both y and r for angle θ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to θ, which is Point P.

    sin(θ)=yr=725
    TIP: Compare the sign of your answer to the CAST diagram. Point P is in Quadrant II. And the CAST diagram tells us that the sine ratio in Quadrant II is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(θ) is 725.


    Submit your answer as:

The trigonometric ratios on the Cartesian plane

On the Cartesian plane the trigonometric ratios are defined in terms of the coordinates x and y, and the radius r. Complete the following equation to make it true (identify the value which belongs in place of the ?):

tanθ=?x
Answer: The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of tangent on the Cartesian plane.


STEP: Use the definition of tangent on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

We use the opposite, adjacent, and hypotenuse sides of a right-angled triangle to define the trigonometric ratios. But we can translate these definitions onto the Cartesian plane using a right-angled triangle on the Cartesian plane.

The diagram below shows a point on the circle at (x;y). The radius of the circle is r. The point makes an angle θ with the positive x-axis. The sides of the triangle have lengths x, y, and r.

Here are three key relationships in the triangle above:

  • The hypotenuse of the triangle is the radius of the circle.
  • The side of the triangle opposite to θ has a length y.
  • The side of the triangle adjacent to θ has a length x.

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

sinθ=oppositehypotenuseyrcosθ=adjacenthypotenusexrtanθ=oppositeadjacentyx

By comparing these to the equation in the question we can find the missing value.

The correct definition is tanθ=yx so the missing quantity is y.


Submit your answer as:

The trigonometric ratios on the Cartesian plane

On the Cartesian plane the trigonometric ratios are defined in terms of the coordinates x and y, and the radius r. Complete the following equation to make it true (identify the value which belongs in place of the ?):

sinθ=?r
Answer: The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of sine on the Cartesian plane.


STEP: Use the definition of sine on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

We use the opposite, adjacent, and hypotenuse sides of a right-angled triangle to define the trigonometric ratios. But we can translate these definitions onto the Cartesian plane using a right-angled triangle on the Cartesian plane.

The diagram below shows a point on the circle at (x;y). The radius of the circle is r. The point makes an angle θ with the positive x-axis. The sides of the triangle have lengths x, y, and r.

Here are three key relationships in the triangle above:

  • The hypotenuse of the triangle is the radius of the circle.
  • The side of the triangle opposite to θ has a length y.
  • The side of the triangle adjacent to θ has a length x.

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

sinθ=oppositehypotenuseyrcosθ=adjacenthypotenusexrtanθ=oppositeadjacentyx

By comparing these to the equation in the question we can find the missing value.

The correct definition is sinθ=yr so the missing quantity is y.


Submit your answer as:

The trigonometric ratios on the Cartesian plane

On the Cartesian plane the trigonometric ratios are defined in terms of the coordinates x and y, and the radius r. Complete the following equation to make it true (identify the value which belongs in place of the ?):

cosθ=x?
Answer: The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of cosine on the Cartesian plane.


STEP: Use the definition of cosine on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

We use the opposite, adjacent, and hypotenuse sides of a right-angled triangle to define the trigonometric ratios. But we can translate these definitions onto the Cartesian plane using a right-angled triangle on the Cartesian plane.

The diagram below shows a point on the circle at (x;y). The radius of the circle is r. The point makes an angle θ with the positive x-axis. The sides of the triangle have lengths x, y, and r.

Here are three key relationships in the triangle above:

  • The hypotenuse of the triangle is the radius of the circle.
  • The side of the triangle opposite to θ has a length y.
  • The side of the triangle adjacent to θ has a length x.

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

sinθ=oppositehypotenuseyrcosθ=adjacenthypotenusexrtanθ=oppositeadjacentyx

By comparing these to the equation in the question we can find the missing value.

The correct definition is cosθ=xr so the missing quantity is r.


Submit your answer as:

Reading ratios from the Cartesian plane

The diagram below shows Point P at (4;3). A triangle is also shown. The lengths of the triangle's sides are 4, 3, and 5, as labelled. This triangle contains an angle which is labelled θ. The angle α is also labelled between the positive x-axis around to the hypotenuse of the triangle. Answer the two questions which follow.

  1. Based on the diagram, what is the value of cosθ?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: cosθ=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start with the fact that the cosine ratio is always the adjacent over the hypotenuse. You can find those values on the triangle in the figure.


    STEP: Read the answer from the triangle in the figure
    [−1 point ⇒ 0 / 1 points left]

    We need to find the value of cosθ. The cosine ratio is always the adjacent over the hypotenuse. So we can find the answer from the labels on the triangle.

    Now write down the ratio:

    cosθ=45
    NOTE: For the angle θ, we needed only the labels in the triangle. This will be different in Question 2.

    The value of cosθ is 45.


    Submit your answer as:
  2. What is the value of cosα?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: cosα= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the cosine ratio on the Cartesian plane.


    STEP: Write down the definition of the cosine ratio on the Cartesian plane
    [−1 point ⇒ 1 / 2 points left]

    In Question 1 we were working with the angle θ. But now we are working with the angle α. So we need to use the definition of cosine on the Cartesian plane. This is for two reasons:

    • Unlike θ, the angle α is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
    • The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive x-axis, and that is where the angle α starts.

    The definition for the cosine ratio is:

    cosα=xr

    STEP: Substitute the values for x and r
    [−1 point ⇒ 0 / 2 points left]

    To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the x-coordinate of Point P, which is 4.

    cosα=xrFor Point P,x=4,r=5cosα=45

    It is now clear that cosα is not the same as cosθ. But the numbers in each ratio are the same: only the signs are different. This is because the angles are closely related: they are both connected to Point P. θ is called a reference angle for α. And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle α.

    NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for cosα depends on which quadrant the reference triangle is in (the size of α). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the cosine ratio is negative in the third quadrant, which tells us that cosα must be negative.

    The value of cosα is 45.


    Submit your answer as:

Reading ratios from the Cartesian plane

The diagram below shows Point P at (4;3). A triangle is also shown. The lengths of the triangle's sides are 4, 3, and 5, as labelled. This triangle contains an angle which is labelled θ. The angle α is also labelled between the positive x-axis around to the hypotenuse of the triangle. Answer the two questions which follow.

  1. Based on the diagram, what is the value of tanθ?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: tanθ=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start with the fact that the tangent ratio is always the opposite over the adjacent. You can find those values on the triangle in the figure.


    STEP: Read the answer from the triangle in the figure
    [−1 point ⇒ 0 / 1 points left]

    We need to find the value of tanθ. The tangent ratio is always the opposite over the adjacent. So we can find the answer from the labels on the triangle.

    Now write down the ratio:

    tanθ=34
    NOTE: For the angle θ, we needed only the labels in the triangle. This will be different in Question 2.

    The value of tanθ is 34.


    Submit your answer as:
  2. What is the value of tanα?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: tanα= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the tangent ratio on the Cartesian plane.


    STEP: Write down the definition of the tangent ratio on the Cartesian plane
    [−1 point ⇒ 1 / 2 points left]

    In Question 1 we were working with the angle θ. But now we are working with the angle α. So we need to use the definition of tangent on the Cartesian plane. This is for two reasons:

    • Unlike θ, the angle α is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
    • The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive x-axis, and that is where the angle α starts.

    The definition for the tangent ratio is:

    tanα=yx

    STEP: Substitute the values for y and x
    [−1 point ⇒ 0 / 2 points left]

    To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the coordinates of Point P, x and y. Note that the x-coordinate is negative and the y-coordinate is positive.

    tanα=yxFor Point P,y=3,x=4tanα=34

    It is now clear that tanα is not the same as tanθ. But the numbers in each ratio are the same: only the signs are different. This is because the angles are closely related: they are both connected to Point P. θ is called a reference angle for α. And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle α.

    NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for tanα depends on which quadrant the reference triangle is in (the size of α). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the tangent ratio is negative in the second quadrant, which tells us that tanα must be negative.

    The value of tanα is 34.


    Submit your answer as:

Reading ratios from the Cartesian plane

The diagram below shows Point P at (12;5). A triangle is also shown. The lengths of the triangle's sides are 12, 5, and 13, as labelled. This triangle contains an angle which is labelled θ. The angle α is also labelled between the positive x-axis around to the hypotenuse of the triangle. Answer the two questions which follow.

  1. Based on the diagram, what is the value of tanθ?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: tanθ=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start with the fact that the tangent ratio is always the opposite over the adjacent. You can find those values on the triangle in the figure.


    STEP: Read the answer from the triangle in the figure
    [−1 point ⇒ 0 / 1 points left]

    We need to find the value of tanθ. The tangent ratio is always the opposite over the adjacent. So we can find the answer from the labels on the triangle.

    Now write down the ratio:

    tanθ=512
    NOTE: For the angle θ, we needed only the labels in the triangle. This will be different in Question 2.

    The value of tanθ is 512.


    Submit your answer as:
  2. What is the value of tanα?

    INSTRUCTION: Give your answer in the form of a fraction.
    Answer: tanα= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the tangent ratio on the Cartesian plane.


    STEP: Write down the definition of the tangent ratio on the Cartesian plane
    [−1 point ⇒ 1 / 2 points left]

    In Question 1 we were working with the angle θ. But now we are working with the angle α. So we need to use the definition of tangent on the Cartesian plane. This is for two reasons:

    • Unlike θ, the angle α is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
    • The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive x-axis, and that is where the angle α starts.

    The definition for the tangent ratio is:

    tanα=yx

    STEP: Substitute the values for y and x
    [−1 point ⇒ 0 / 2 points left]

    To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the coordinates of Point P, x and y. Note that the x-coordinate is negative and the y-coordinate is negative.

    tanα=yxFor Point P,y=5,x=12tanα=512=512

    It is now clear that tanα is the same as tanθ. In this calculation both parts of the ratio were negative, but the ratio ends up positive. That means we get the same answer as in Question 1. This is because the angles are closely related: they are both connected to Point P. θ is called a reference angle for α. And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle α.

    NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for tanα depends on which quadrant the reference triangle is in (the size of α). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the tangent ratio is positive in the third quadrant, which tells us that tanα must be positive.

    The value of tanα is 512.


    Submit your answer as:

Simplifying trigonometric expressions with reduction formulas

Determine the value of the following without using a calculator.

sin(150°)
INSTRUCTIONS:
  • Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
  • Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.
Answer: sin(150°)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of 90°,180°, or 360°. The other part should be one of the special angles, 30°,45°, or 60°.


STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

  • We must show the working step by step.
  • We can expect to use special angles to solve the problem. The special angles are 30°,45°, and 60°.

The first step is to break the angle in the question into two terms so that we can use one of the reduction formulas. We want one of those pieces to be a special angle because that is the only way we can evaluate the expression without a calculator. In this case, the original angle is 150°. We can split that into the values 90° and 60° using addition.

sin(150°)=sin(90°+60°)

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. In this case the angle is 90°. So we will use a co-function identity. That means sin changes to cos:

sin(90°+θ)=cosθ
TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression sin(90°+60°). Note that there is a negative at the front of the expression. But it is not part of the reduction formula, so it will just sit there until the end.

sin(90°+60°)=cos(60°)

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

=cos(60°)=12

The final answer is sin(150°)=12.


Submit your answer as:

Simplifying trigonometric expressions with reduction formulas

Without using a calculator, determine the value of the following expression.

sin(210°)
INSTRUCTIONS:
  • Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
  • Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.
Answer: sin(210°)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of 90°,180°, or 360°. The other part should be one of the special angles, 30°,45°, or 60°.


STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

  • We must show the working step by step.
  • We can expect to use special angles to solve the problem. The special angles are 30°,45°, and 60°.

The first step is to break the angle in the question into two terms so that we can use one of the reduction formulas. We want one of those pieces to be a special angle because that is the only way we can evaluate the expression without a calculator. In this case, the original angle is 210°. We can split that into the values 180° and 30° using addition.

sin(210°)=sin(180°+30°)

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. Specifically, we can remove the 180° part of the angle using the formula:

sin(180°+θ)=sinθ
TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression sin(180°+30°). Note that there is a negative at the front of the expression. But the reduction formula above shows another negative will come into our solution. So we will find two negatives in the first step below.

sin(180°+30°)=(sin(30°))=sin(30°)

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

=sin(30°)=12

The final answer is sin(210°)=12.


Submit your answer as:

Simplifying trigonometric expressions with reduction formulas

Without using a calculator, determine the value of the following expression.

sin(315°)
INSTRUCTIONS:
  • Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
  • Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.
Answer: sin(315°)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of 90°,180°, or 360°. The other part should be one of the special angles, 30°,45°, or 60°.


STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

  • We must show the working step by step.
  • We can expect to use special angles to solve the problem. The special angles are 30°,45°, and 60°.

The first step is to break the angle in the question into two terms so that we can use one of the reduction formulas. We want one of those pieces to be a special angle because that is the only way we can evaluate the expression without a calculator. In this case, the original angle is 315°. We can split that into the values 360° and 45° using subtraction.

sin(315°)=sin(360°45°)

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. Specifically, we can remove the 360° part of the angle using the formula:

sin(360°θ)=sinθ
TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression sin(360°45°). Note that the reduction formula has a negative on the right side. So we will pick up a negative in the work below.

sin(360°45°)=sin(45°)

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

=sin(45°)=22

The final answer is sin(315°)=22.


Submit your answer as:

Special angles: 0° and 90°

  1. On the Cartesian plane, the cosine ratio is defined as xr. Based on this definition or otherwise, determine the value of cos0°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of cos0° is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of cos0°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate xr
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of cos0° is 1.


    Submit your answer as:
  2. Question 1 was about the value of cos0°. Which of the reasons below explains the answer to Question 1?

    Answer:

    The explanation for the answer to Question 1 is that when θ=0°, .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=0°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why cos0° is 1. The answer comes directly from the definition for the cosine ratio on the Cartesian plane. That definition is:

    cosθ=xr

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 0°. When that happens, the hypotenuse will rotate down. And the triangle gets more flat until the hypotenuse falls onto the horizontal side of the triangle on the x-axis.

    For θ=0°, the triangle collapses to a horizontal line segment on the x-axis. The segment reaches from the origin over to the circle at the point (x;0). And the following things are both true about the "triangle" represented by that line:

    • x=r
    • y=0

    Let's use this information in the definition of the cosine ratio on the Cartesian plane:

    cosθ=xrcos0°=rr=1

    And finally we can see why cos0° must be 1: it is because when the angle is zero, the side of the triangle corresponding to x is the same length as the radius.

    The reason cos0° is 1 is because x=r.


    Submit your answer as:

Special angles: 0° and 90°

  1. On the Cartesian plane, the tangent ratio is defined as yx. Based on this definition or otherwise, determine the value of tan90°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of tan90° is .
    string
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of tan90°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate yx
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of tan90° is undefined.


    Submit your answer as:
  2. Question 1 was about the value of tan90°. Which of the reasons below explains the answer to Question 1?

    Answer:

    The explanation for the answer to Question 1 is that when θ=90°, .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=90°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why tan90° is undefined. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

    tanθ=yx

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 90°. When that happens, the hypotenuse will rotate up. And the triangle gets more thin until the hypotenuse and the vertical side swing up and over to meet on the y-axis.

    For θ=90°, the triangle collapses to a vertical line segment on the y-axis. The segment reaches from the origin up to the circle at the point (0;y). And the following things are both true about the "triangle" represented by that line:

    • x=0
    • y=r

    Let's use this information in the definition of the tangent ratio on the Cartesian plane:

    tanθ=yxtan90°=y0=undefined

    And finally we can see why tan90° must be undefined: it is because when the angle is 90°, the side of the triangle corresponding to x shrinks to zero, leading to division by zero.

    The reason tan90° is undefined is because x=0.


    Submit your answer as:

Special angles: 0° and 90°

  1. On the Cartesian plane, the sine ratio is defined as yr. Based on this definition or otherwise, determine the value of sin0°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of sin0° is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of sin0°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate yr
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of sin0° is 0.


    Submit your answer as:
  2. Question 1 was about the value of sin0°. Which of the reasons below explains the answer to Question 1?

    Answer:

    The explanation for the answer to Question 1 is that when θ=0°, .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=0°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why sin0° is 0. The answer comes directly from the definition for the sine ratio on the Cartesian plane. That definition is:

    sinθ=yr

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 0°. When that happens, the hypotenuse will rotate down. And the triangle gets more flat until the hypotenuse falls onto the horizontal side of the triangle on the x-axis.

    For θ=0°, the triangle collapses to a horizontal line segment on the x-axis. The segment reaches from the origin over to the circle at the point (x;0). And the following things are both true about the "triangle" represented by that line:

    • x=r
    • y=0

    Let's use this information in the definition of the sine ratio on the Cartesian plane:

    sinθ=yrsin0°=0r=0

    And finally we can see why sin0° must be 0: it is because when the angle is zero, the side of the triangle corresponding to y shrinks down to zero.

    The reason sin0° is 0 is because y=0.


    Submit your answer as:

Special angles: 180°, 270°, and 360°

  1. On the Cartesian plane, the sine function is defined as yr. Based on this definition or otherwise, determine the value of sin180°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of sin180° is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of sin180°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate yr
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of sin180° is 0.


    Submit your answer as:
  2. Question 1 was about the value of sin180°. Which of the reasons below explains the answer to Question 1?

    Answer: The explanation for the answer to Question 1 is that when θ=180°, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=180°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why sin180° is 0. The answer comes directly from the definition for the sine ratio on the Cartesian plane. That definition is:

    sinθ=yr

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 180°. When that happens, the hypotenuse will rotate past the y-axis into Quadrant II, and around until it reaches the negative x-axis (pointing to the left). This is shown below.

    For θ=180°, the triangle collapses to a horizontal line segment on the x-axis. The segment reaches from the origin over to the circle at the point (x;0). And the following things are both true about the "triangle" represented by that line:

    • x=r
    • y=0

    Remember that r, the radius, is positive. But for θ=180° we know that x must be negative (you can see this on the diagram above). That is the reason why x=r, not x=r.

    Now we can use this information in the definition of the sine ratio on the Cartesian plane:

    sinθ=yrsin180°=0r=0

    And finally we can see why sin180° must be 0: it is because when the angle is 180°, the side of the triangle corresponding to y shrinks down to zero.

    The reason sin180° is 0 is because y=0.


    Submit your answer as:

Special angles: 180°, 270°, and 360°

  1. On the Cartesian plane, the tangent function is defined as yx. Based on this definition or otherwise, determine the value of tan180°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of tan180° is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of tan180°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate yx
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of tan180° is 0.


    Submit your answer as:
  2. Question 1 was about the value of tan180°. Which of the reasons below explains the answer to Question 1?

    Answer: The explanation for the answer to Question 1 is that when θ=180°, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=180°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why tan180° is 0. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

    tanθ=yx

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 180°. When that happens, the hypotenuse will rotate past the y-axis into Quadrant II, and around until it reaches the negative x-axis (pointing to the left). This is shown below.

    For θ=180°, the triangle collapses to a horizontal line segment on the x-axis. The segment reaches from the origin over to the circle at the point (x;0). And the following things are both true about the "triangle" represented by that line:

    • x=r
    • y=0

    Remember that r, the radius, is positive. But for θ=180° we know that x must be negative (you can see this on the diagram above). That is the reason why x=r, not x=r.

    Now we can use this information in the definition of the tangent ratio on the Cartesian plane:

    tanθ=yxtan180°=0x=0

    And finally we can see why tan180° must be 0: it is because when the angle is 180°, the side of the triangle corresponding to y shrinks down to zero.

    The reason tan180° is 0 is because y=0.


    Submit your answer as:

Special angles: 180°, 270°, and 360°

  1. On the Cartesian plane, the tangent function is defined as yx. Based on this definition or otherwise, determine the value of tan180°.

    INSTRUCTION:
    • You may use a calculator.
    • If the answer is undefined, type undefined.
    Answer: The value of tan180° is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!


    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the value of tan180°. Here are two ways you might find the answer:

    • draw a triangle on the Cartesian plane to evaluate yx
    • use a calculator

    The reason for the answer is the subject of Question 2, below.

    The value of tan180° is 0.


    Submit your answer as:
  2. Question 1 was about the value of tan180°. Which of the reasons below explains the answer to Question 1?

    Answer: The explanation for the answer to Question 1 is that when θ=180°, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use a picture of a triangle on the Cartesian plane with θ=180°.


    STEP: Use a triangle on the Cartesian plane to answer the question
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why tan180° is 0. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

    tanθ=yx

    Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

    The angle θ starts at the positive x-axis and curves up to meet the hypotenuse of the triangle. Imagine that θ becomes 180°. When that happens, the hypotenuse will rotate past the y-axis into Quadrant II, and around until it reaches the negative x-axis (pointing to the left). This is shown below.

    For θ=180°, the triangle collapses to a horizontal line segment on the x-axis. The segment reaches from the origin over to the circle at the point (x;0). And the following things are both true about the "triangle" represented by that line:

    • x=r
    • y=0

    Remember that r, the radius, is positive. But for θ=180° we know that x must be negative (you can see this on the diagram above). That is the reason why x=r, not x=r.

    Now we can use this information in the definition of the tangent ratio on the Cartesian plane:

    tanθ=yxtan180°=0x=0

    And finally we can see why tan180° must be 0: it is because when the angle is 180°, the side of the triangle corresponding to y shrinks down to zero.

    The reason tan180° is 0 is because y=0.


    Submit your answer as:

Trigonometry gone wild!

  1. Point K is given on the Cartesian plane with the origin at O. It is in the fourth quadrant at (18;yK), where yK is the y-coordinate of Point K. K makes an angle ϕ with the positive x-axis, as labelled, and OK¯ is 30 units long. Given that tanϕ=43, determine the value of yK. The diagram may or may not be drawn to scale.

    Answer: yK =
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the tangent ratio on the Cartesian plane to write an equation which includes the y-coordinate.


    STEP: Write an equation based on the information given
    [−1 point ⇒ 1 / 2 points left]

    We need to determine the y-coordinate of Point K. The point sits at an angle of ϕ to the positive x-axis and we know that tanϕ is equal to 43. As always, the tangent ratio is defined as yx. So we can write:

    yx=43

    This equation is true for the angle ϕ and Point K. So we can substitute in the x-coordinate value of 18 on the left-hand side of the equation. (We will also change y to yK because the equation is specifically about Point K now.)

    yK(18)=43
    NOTE:

    The equation above is a proportion. It represents the fact that any point at an angle of ϕ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle ϕ. The second point makes a smaller triangle.

    The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!


    STEP: Solve the equation
    [−1 point ⇒ 0 / 2 points left]

    Now we need to solve the equation. Multiply both sides by the denominator on the left side to isolate yK.

    yK(18)=43yK=43(18)yK=24
    TIP: Check the sign of the answer: it must be negative because Point K is in the fourth quadrant. If your sign is wrong, you should go back and check through your work.

    The y-coordinate of Point K is −24.


    Submit your answer as:
  2. Evaluate the following expression:

    5cos2ϕ9
    INSTRUCTION: Your answer must be a simplified fraction.
    Answer: 5cos2ϕ9=
    fraction
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the result of Question 1 to calculate the value of cosϕ. Then evaluate the expression.


    STEP: Determine the value of cosϕ
    [−1 point ⇒ 1 / 2 points left]

    We can start by finding the value of cosϕ. From Question 1 we know everything about Point K (which is linked to the angle ϕ):

    On the Cartesian plane the cosine ratio is xr. Using the values from the figure above we get:

    cosθ=xrcosϕ=(18)(30)=35

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now we can evaluate the expression. Substitute 35 in for cosϕ and simplify:

    5cos2ϕ9=5(35)29=5(925)9=959=95455=545

    The value of the expression is 545.


    Submit your answer as:

Trigonometry gone wild!

  1. Point K is given on the Cartesian plane with the origin at O. It is in the second quadrant at (30;yK), where yK is the y-coordinate of Point K. K makes an angle ϕ with the positive x-axis, as labelled, and OK¯ is 34 units long. Given that sinϕ=817, determine the value of yK. The diagram may or may not be drawn to scale.

    Answer: yK =
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the sine ratio on the Cartesian plane to write an equation which includes the y-coordinate.


    STEP: Write an equation based on the information given
    [−1 point ⇒ 1 / 2 points left]

    We need to determine the y-coordinate of Point K. The point sits at an angle of ϕ to the positive x-axis and we know that sinϕ is equal to 817. As always, the sine ratio is defined as yr. So we can write:

    yr=817

    This equation is true for the angle ϕ and Point K. So we can substitute in the radius value of 34 on the left-hand side of the equation. (We will also change y to yK because the equation is specifically about Point K now.)

    yK(34)=817
    NOTE:

    The equation above is a proportion. It represents the fact that any point at an angle of ϕ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle ϕ. The second point makes a smaller triangle.

    The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!


    STEP: Solve the equation
    [−1 point ⇒ 0 / 2 points left]

    Now we need to solve the equation. Multiply both sides by the denominator on the left side to isolate yK.

    yK(34)=817yK=817(34)yK=16
    TIP: Check the sign of the answer: it must be positive because Point K is in the second quadrant. If your sign is wrong, you should go back and check through your work.

    The y-coordinate of Point K is 16.


    Submit your answer as:
  2. Evaluate the following expression:

    3445cos2ϕ+4
    INSTRUCTION: Your answer must be a simplified fraction.
    Answer: 3445cos2ϕ+4=
    fraction
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the result of Question 1 to calculate the value of cosϕ. Then evaluate the expression.


    STEP: Determine the value of cosϕ
    [−1 point ⇒ 1 / 2 points left]

    We can start by finding the value of cosϕ. From Question 1 we know everything about Point K (which is linked to the angle ϕ):

    On the Cartesian plane the cosine ratio is xr. Using the values from the figure above we get:

    cosθ=xrcosϕ=(30)(34)=1517

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now we can evaluate the expression. Substitute 1517 in for cosϕ and simplify:

    3445cos2ϕ+4=3445(1517)2+4=3445(225289)+4=1017+4=1017+6817=5817

    The value of the expression is 5817.


    Submit your answer as:

Trigonometry gone wild!

  1. Point M is given on the Cartesian plane with the origin at O. It is in the fourth quadrant at (xM;15), where xM is the x-coordinate of Point M. M makes an angle ϕ with the positive x-axis, as labelled, and OM¯ is 39 units long. Given that cosϕ=1213, determine the value of xM. The diagram may or may not be drawn to scale.

    Answer: xM =
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition of the cosine ratio on the Cartesian plane to write an equation which includes the x-coordinate.


    STEP: Write an equation based on the information given
    [−1 point ⇒ 1 / 2 points left]

    We need to determine the x-coordinate of Point M. The point sits at an angle of ϕ to the positive x-axis and we know that cosϕ is equal to 1213. As always, the cosine ratio is defined as xr. So we can write:

    xr=1213

    This equation is true for the angle ϕ and Point M. So we can substitute in the radius value of 39 on the left-hand side of the equation. (We will also change x to xM because the equation is specifically about Point M now.)

    xM(39)=1213
    NOTE:

    The equation above is a proportion. It represents the fact that any point at an angle of ϕ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle ϕ. The second point makes a smaller triangle.

    The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!


    STEP: Solve the equation
    [−1 point ⇒ 0 / 2 points left]

    Now we need to solve the equation. Multiply both sides by the denominator on the left side to isolate xM.

    xM(39)=1213xM=1213(39)xM=36
    TIP: Check the sign of the answer: it must be positive because Point M is in the fourth quadrant. If your sign is wrong, you should go back and check through your work.

    The x-coordinate of Point M is 36.


    Submit your answer as:
  2. Evaluate the following expression:

    12tan2ϕ+5
    INSTRUCTION: Your answer must be a simplified fraction.
    Answer: 12tan2ϕ+5=
    fraction
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the result of Question 1 to calculate the value of tanϕ. Then evaluate the expression.


    STEP: Determine the value of tanϕ
    [−1 point ⇒ 1 / 2 points left]

    We can start by finding the value of tanϕ. From Question 1 we know everything about Point M (which is linked to the angle ϕ):

    On the Cartesian plane the tangent ratio is yx. Using the values from the figure above we get:

    tanθ=yxtanϕ=(15)(36)=512

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now we can evaluate the expression. Substitute 512 in for tanϕ and simplify:

    12tan2ϕ+5=12(512)2+5=12(25144)+5=2512+5=2512+6012=3512

    The value of the expression is 3512.


    Submit your answer as:

Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter T belong?

Answer:

The T should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

The letters sit on the Cartesian plane as shown below: C is in the lower right quadrant (Quadrant IV). The letters C-A-S-T go around the diagram in the anticlockwise direction. You will find it helpful to memorise the positions of the letters!

The T belongs in Quadrant III. The T in Quadrant III tells us that the tangent ratio is positive for any angle in Quadrant III. At the same time, it tells us that the sine and cosine ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The T should be in Quadrant III.


Submit your answer as:

Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter A belong?

Answer:

The A should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

The letters sit on the Cartesian plane as shown below: C is in the lower right quadrant (Quadrant IV). The letters C-A-S-T go around the diagram in the anticlockwise direction. You will find it helpful to memorise the positions of the letters!

The A belongs in Quadrant I. The A in Quadrant I tells us that all three of the trigonometric ratios are positive for any angle in Quadrant I (any angle between 0° and 90°).

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the sine ratio is positive in Quadrants I and II, while it is negative in Quadrants III and IV.

The A should be in Quadrant I.


Submit your answer as:

Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter T belong?

Answer:

The T should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

The letters sit on the Cartesian plane as shown below: C is in the lower right quadrant (Quadrant IV). The letters C-A-S-T go around the diagram in the anticlockwise direction. You will find it helpful to memorise the positions of the letters!

The T belongs in Quadrant III. The T in Quadrant III tells us that the tangent ratio is positive for any angle in Quadrant III. At the same time, it tells us that the sine and cosine ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the tangent ratio is positive in Quadrants I and III, while it is negative in Quadrants II and IV.

The T should be in Quadrant III.


Submit your answer as:

Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle θ:

cosθ=1517sinθ=817

We can draw a reference triangle on the Cartesian plane to represent the angle θ. One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).
Answer: The coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the cosine ratio is positive and the second equation shows that the sine ratio is negative. Use these two facts and the CAST diagram to figure out which quadrant the angle θ must point into.


STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We need to determine the coordinates of the point on the Cartesian plane which correspond to the reference triangle for the angle θ. The first thing we need to do is figure out which quadrant the angle points into. Then we can draw a point and the reference triangle we need in that quadrant.

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle θ must point into. The first equation shows that the cosine ratio is positive and the second equation shows that the sine ratio is negative. So the point must be in Quadrant IV: that is the only quadrant where the cosine ratio is positive and the sine ratio is negative.


STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant IV, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant IV.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of x and y from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the cosine and sine ratios.

The definition for cosine is xr. The x in this definition is one of the coordinates we want!

cosθ=1517xrwhich means:x=15r=17

The definition for sine is yr. And we can follow the same logic as above to find the value of y. In this case, the ratio value is negative. That means the The y-value must be negative because the radius is always positive.

sinθ=817yrwhich means:y=8r=17

Now we know that the coordinates of the point in the reference triangle are (15;8). And we also know the radius, r, for the reference triangle. The reference triangle looks like this:

NOTE: While the lengths of the triangle's sides must be positive, the coordinates of the point can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

The coordinates of the point in the reference triangle are (15;8).


Submit your answer as:

Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle θ:

cosθ=35sinθ=45

We can draw a reference triangle on the Cartesian plane to represent the angle θ. One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).
Answer: The coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the cosine ratio is negative and the second equation shows that the sine ratio is positive. Use these two facts and the CAST diagram to figure out which quadrant the angle θ must point into.


STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We need to determine the coordinates of the point on the Cartesian plane which correspond to the reference triangle for the angle θ. The first thing we need to do is figure out which quadrant the angle points into. Then we can draw a point and the reference triangle we need in that quadrant.

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle θ must point into. The first equation shows that the cosine ratio is negative and the second equation shows that the sine ratio is positive. So the point must be in Quadrant II: that is the only quadrant where the cosine ratio is negative and the sine ratio is positive.


STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant II, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant II.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of x and y from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the cosine and sine ratios.

The definition for cosine is xr. The x in this definition is one of the coordinates we want! In this case, the ratio value is negative. That means the the x-value must be negative because the radius is always positive.

cosθ=35xrwhich means:x=3r=5

The definition for sine is yr. And we can follow the same logic as above to find the value of y.

sinθ=45yrwhich means:y=4r=5

Now we know that the coordinates of the point in the reference triangle are (3;4). And we also know the radius, r, for the reference triangle. The reference triangle looks like this:

NOTE: While the lengths of the triangle's sides must be positive, the coordinates of the point can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

The coordinates of the point in the reference triangle are (3;4).


Submit your answer as:

Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle θ:

cosθ=817tanθ=158

We can draw a reference triangle on the Cartesian plane to represent the angle θ. One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).
Answer: The coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the cosine ratio is negative and the second equation shows that the tangent ratio is positive. Use these two facts and the CAST diagram to figure out which quadrant the angle θ must point into.


STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We need to determine the coordinates of the point on the Cartesian plane which correspond to the reference triangle for the angle θ. The first thing we need to do is figure out which quadrant the angle points into. Then we can draw a point and the reference triangle we need in that quadrant.

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle θ must point into. The first equation shows that the cosine ratio is negative and the second equation shows that the tangent ratio is positive. So the point must be in Quadrant III: that is the only quadrant where the cosine ratio is negative and the tangent ratio is positive.


STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant III, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant III.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of x and y from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the cosine and tangent ratios.

In fact, the tangent ratio is enough to tell us the coordinates. This is because the tangent ratio is always equal to yx. So we can pull the values of x and y out of this equation:

tanθ=158yx

The ratio is positive. But that does not mean that x and y are both positive: they might both be negative (because a negative divided by a negative makes a positive). So either x and y are both positive, or they are both negative. Which is it? The answer comes from the fact that we already know which quadrant the point is in! The point is in Quadrant III, which means both of the coordinates must be negative. Therefore:

x=8y=15

Now we know that the coordinates of the point in the reference triangle are (8;15). Note that you can also read the radius value (the hypotenuse of the triangle) from the cosine ratio, which is equal to 817. The radius is equal to 17. The reference triangle looks like this:

NOTE: While the lengths of the triangle's sides must be positive, the coordinates of the point can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

The coordinates of the point in the reference triangle are (8;15).


Submit your answer as:

Trigonometric ratios on the Cartesian plane: positives & negatives

  1. Consider the following equation about an angle θ:

    cosθ=45

    The diagram below shows four points on the Cartesian plane. They are labelled W,X,Y, and Z. The diagram is not drawn to scale.

    Two of the four points above correspond to the equation cosθ=45. Which two?

    Answer:

    The points corresponding to cosθ=45 are Points .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to compare the sign of the ratio to the CAST diagram.


    STEP: Use the CAST diagram and the sign of the ratio given
    [−1 point ⇒ 0 / 1 points left]

    On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

    In this case we know that:

    cosθ=45positive

    This equation tells us that the angle θ must point into one of the two quadrants where the cosine ratio is positive. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

    We are dealing with the cosine ratio, and we know that the ratio has a positive value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. And the C in Quadrant IV means that the cosine ratio is positive in that quadrant. So Points W and Z are the points which correspond to the equation in the question.

    The correct points are W and Z.


    Submit your answer as:
  2. In Question 1 you found the two points which correspond to this equation:

    cosθ=45

    There are two values for θ connected to those points. From the list of choices below, select the two values closest to θ (the choices in the list have been rounded).

    TIP: Use the answer to Question 1.
    Answer: The values for θ are closest to: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to pick the angles which match the correct points in Question 1.


    STEP: Use the answer to Question 1
    [−1 point ⇒ 0 / 1 points left]

    For this question we need to identify the values of θ which correspond to the equation

    cosθ=45

    There are two such values: they match the points we identified in Question 1.

    We already know that Points W and Z correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: 37°;143°;217°;323°. Two of these choices agree with the positions of Points W and Z. (Remember that the angles always start on the positive x-axis, as shown below.)

    We do not know the exact values of these two angles. But we know that one of them points into Quadrant I and the other points into Quadrant IV. That means the smaller angle must be acute (between 0° and 90°) and the larger angle must be between 270° and 360°. Based on the choices in the list, the values for θ must be 37° and 323°.

    The two values for θ are 37° and 323°.


    Submit your answer as:

Trigonometric ratios on the Cartesian plane: positives & negatives

  1. Consider the following equation about an angle θ:

    tanθ=43

    The diagram below shows four points on the Cartesian plane. They are labelled W,X,Y, and Z. The diagram is not drawn to scale.

    Two of the four points above correspond to the equation tanθ=43. Which two?

    Answer:

    The points corresponding to tanθ=43 are Points .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to compare the sign of the ratio to the CAST diagram.


    STEP: Use the CAST diagram and the sign of the ratio given
    [−1 point ⇒ 0 / 1 points left]

    On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

    In this case we know that:

    tanθ=43positive

    This equation tells us that the angle θ must point into one of the two quadrants where the tangent ratio is positive. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

    We are dealing with the tangent ratio, and we know that the ratio has a positive value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. And the T in Quadrant III means that the tangent ratio is positive in that quadrant. So Points W and Y are the points which correspond to the equation in the question.

    The correct points are W and Y.


    Submit your answer as:
  2. In Question 1 you found the two points which correspond to this equation:

    tanθ=43

    There are two values for θ connected to those points. From the list of choices below, select the two values closest to θ (the choices in the list have been rounded).

    TIP: Use the answer to Question 1.
    Answer: The values for θ are closest to: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to pick the angles which match the correct points in Question 1.


    STEP: Use the answer to Question 1
    [−1 point ⇒ 0 / 1 points left]

    For this question we need to identify the values of θ which correspond to the equation

    tanθ=43

    There are two such values: they match the points we identified in Question 1.

    We already know that Points W and Y correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: 53°;127°;233°;307°. Two of these choices agree with the positions of Points W and Y. (Remember that the angles always start on the positive x-axis, as shown below.)

    We do not know the exact values of these two angles. But we know that one of them points into Quadrant I and the other points into Quadrant III. That means the smaller angle must be acute (between 0° and 90°) and the larger angle must be between 180° and 270°. Based on the choices in the list, the values for θ must be 53° and 233°.

    The two values for θ are 53° and 233°.


    Submit your answer as:

Trigonometric ratios on the Cartesian plane: positives & negatives

  1. Consider the following equation about an angle θ:

    cosθ=1517

    The diagram below shows four points on the Cartesian plane. They are labelled W,X,Y, and Z. The diagram is not drawn to scale.

    Two of the four points above correspond to the equation cosθ=1517. Which two?

    Answer:

    The points corresponding to cosθ=1517 are Points .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to compare the sign of the ratio to the CAST diagram.


    STEP: Use the CAST diagram and the sign of the ratio given
    [−1 point ⇒ 0 / 1 points left]

    On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

    In this case we know that:

    cosθ=1517positive

    This equation tells us that the angle θ must point into one of the two quadrants where the cosine ratio is positive. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

    We are dealing with the cosine ratio, and we know that the ratio has a positive value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. And the C in Quadrant IV means that the cosine ratio is positive in that quadrant. So Points W and Z are the points which correspond to the equation in the question.

    The correct points are W and Z.


    Submit your answer as:
  2. In Question 1 you found the two points which correspond to this equation:

    cosθ=1517

    There are two values for θ connected to those points. From the list of choices below, select the two values closest to θ (the choices in the list have been rounded).

    TIP: Use the answer to Question 1.
    Answer: The values for θ are closest to: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to pick the angles which match the correct points in Question 1.


    STEP: Use the answer to Question 1
    [−1 point ⇒ 0 / 1 points left]

    For this question we need to identify the values of θ which correspond to the equation

    cosθ=1517

    There are two such values: they match the points we identified in Question 1.

    We already know that Points W and Z correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: 28°;152°;208°;332°. Two of these choices agree with the positions of Points W and Z. (Remember that the angles always start on the positive x-axis, as shown below.)

    We do not know the exact values of these two angles. But we know that one of them points into Quadrant I and the other points into Quadrant IV. That means the smaller angle must be acute (between 0° and 90°) and the larger angle must be between 270° and 360°. Based on the choices in the list, the values for θ must be 28° and 332°.

    The two values for θ are 28° and 332°.


    Submit your answer as:

Finding a ratio from a point

The figure below shows Point K at (x;12). It is a distance 20 from the origin, as labelled. The angle between the positive x-axis and Point K is B. This diagram is not drawn to scale.

Answer the two question which follow about Point K and angle B.

  1. Find the x-coordinate of Point K.

    Answer: The x-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.


    STEP: Draw a right-angled triangle onto the given figure
    [−1 point ⇒ 2 / 3 points left]

    We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the y-coordinate of Point K.


    STEP: Use the triangle to calculate the missing side
    [−1 point ⇒ 1 / 3 points left]

    Remember that we want to find the x-coordinate of Point K. We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

    a2+b2=c2x2+(12)2=(20)2x2+144=400x2=256x=±256x=±16

    This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is 16 units.


    STEP: Check the sign and adjust it if necessary
    [−1 point ⇒ 0 / 3 points left]

    The result above tells us that Point K is 16 units away from the y-axis. So the missing coordinate must be either 16 or 16. It depends on the position of the point! From the diagram we can see that Point K is in Quadrant II, so the x-coordinate has to be negative.

    NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: 3 spaces right is not the same as 3 spaces left. It is crucial to be aware that these signs can be different.

    The x-coordinate of Point K is x=16.


    Submit your answer as:
  2. Now determine the value of cos(B).

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer: cos(B)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the coordinates of the point and the radius, whichever are necessary.


    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the cosine ratio on the Cartesian plane is:

    cos(θ)=xr

    Point K corresponds to angle B, so all we need to do is substitute in the correct values from the point. In this case that means using the x-coordinate and the radius.

    cos(θ)=xrcos(B)=1620=45
    TIP: Compare the sign of your answer to the CAST diagram. Point K is in Quadrant II. And the CAST diagram tells us that the cosine ratio in Quadrant II is always negative. Make sure the sign of your answer agrees with the CAST diagram.

    The value of cos(B) is 45.


    Submit your answer as:

Finding a ratio from a point

The figure below shows Point K at (x;6). It is a distance 10 from the origin, as labelled. The angle between the positive x-axis and Point K is B. This diagram is not drawn to scale.

Answer the two question which follow about Point K and angle B.

  1. Find the x-coordinate of Point K.

    Answer: The x-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.


    STEP: Draw a right-angled triangle onto the given figure
    [−1 point ⇒ 2 / 3 points left]

    We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the y-coordinate of Point K.

    Notice that the side of the triangle is labelled with a positive number, even though the y-coordinate is negative. See the note below for an explanation about why these signs are not always the same.


    STEP: Use the triangle to calculate the missing side
    [−1 point ⇒ 1 / 3 points left]

    Remember that we want to find the x-coordinate of Point K. We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

    a2+b2=c2x2+(6)2=(10)2x2+36=100x2=64x=±64x=±8

    This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is 8 units.


    STEP: Check the sign and adjust it if necessary
    [−1 point ⇒ 0 / 3 points left]

    The result above tells us that Point K is 8 units away from the y-axis. So the missing coordinate must be either 8 or 8. It depends on the position of the point! From the diagram we can see that Point K is in Quadrant IV, so the x-coordinate has to be positive.

    NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: 3 spaces right is not the same as 3 spaces left. It is crucial to be aware that these signs can be different.

    The x-coordinate of Point K is x=8.


    Submit your answer as:
  2. Now determine the value of cos(B).

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer: cos(B)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the coordinates of the point and the radius, whichever are necessary.


    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the cosine ratio on the Cartesian plane is:

    cos(θ)=xr

    Point K corresponds to angle B, so all we need to do is substitute in the correct values from the point. In this case that means using the x-coordinate and the radius.

    cos(θ)=xrcos(B)=810=45
    TIP: Compare the sign of your answer to the CAST diagram. Point K is in Quadrant IV. And the CAST diagram tells us that the cosine ratio in Quadrant IV is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of cos(B) is 45.


    Submit your answer as:

Finding a ratio from a point

The figure below shows Point M at (x;9). It is a distance 15 from the origin, as labelled. The angle between the positive x-axis and Point M is B. This diagram is not drawn to scale.

Answer the two question which follow about Point M and angle B.

  1. Determine the x-coordinate of Point M.

    Answer: The x-coordinate is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.


    STEP: Draw a right-angled triangle onto the given figure
    [−1 point ⇒ 2 / 3 points left]

    We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the y-coordinate of Point M.


    STEP: Use the triangle to calculate the missing side
    [−1 point ⇒ 1 / 3 points left]

    Remember that we want to find the x-coordinate of Point M. We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

    a2+b2=c2x2+(9)2=(15)2x2+81=225x2=144x=±144x=±12

    This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is 12 units.


    STEP: Check the sign and adjust it if necessary
    [−1 point ⇒ 0 / 3 points left]

    The result above tells us that Point M is 12 units away from the y-axis. So the missing coordinate must be either 12 or 12. It depends on the position of the point! From the diagram we can see that Point M is in Quadrant I, so the x-coordinate has to be positive.

    NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: 3 spaces right is not the same as 3 spaces left. It is crucial to be aware that these signs can be different.

    The x-coordinate of Point M is x=12.


    Submit your answer as:
  2. Now determine the value of tan(B).

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer: tan(B)= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the coordinates of the point and the radius, whichever are necessary.


    STEP: Use the definition of the tangent ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    The definition of the tangent ratio on the Cartesian plane is:

    tan(θ)=yx

    Point M corresponds to angle B, so all we need to do is substitute in the correct values from the point. In this case that means using both x and y from Point P.

    tan(θ)=yxtan(B)=912=34
    TIP: Compare the sign of your answer to the CAST diagram. Point M is in Quadrant I. And the CAST diagram tells us that the tangent ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of tan(B) is 34.


    Submit your answer as:

Finding one ratio from another

Given:

cos(C)=59

Determine the value of tan(C) if C is obtuse.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).
Answer: tan(C)= .
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle C is in and draw a sketch to represent that angle.


STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle C points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is obtuse, the angle must be in Quadrant II. This agrees with the fact that cos(C) is negative. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the cosine ratio is the ratio of x to r. While the value of cos(C) is negative, the labels on the triangle are positive because lengths are always positive.

NOTE: The figure above is a sketch: the triangle is probably not actually the size shown above. That is not a problem, but it means that we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Find the value of tan(C)
[−2 points ⇒ 0 / 4 points left]

We want to find tan(C), which means we want the ratio yx. So we need the value of y. We can find this using the theorem of Pythagoras with the reference triangle:

a2+b2=c2y2+(5)2=(9)2y2=8125y2=56y2=414y2=±414y=±214

This calculation tells us two things: the third side of the triangle is 214 units long, and the y-coordinate of the point in our diagram is 214. Now we have enough information to evaluate the tangent ratio for angle C.

tan(C)=yx=2145=2145
TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant II and we expect the tangent ratio to be negative in this quadrant. So all is groovy!

The value of tan(C) is 2145.


Submit your answer as:

Finding one ratio from another

Given:

sin(C)=711

Determine the value of cos(C) if C is greater than 0° and less than 270°.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).
Answer: cos(C)= .
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle C is in and draw a sketch to represent that angle.


STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle C points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is greater than 0° and less than 270°, and the fact that sin(C) is negative, the angle must be in Quadrant III. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the sine ratio is the ratio of y to r. While the value of sin(C) is negative, the labels on the triangle are positive because lengths are always positive.

NOTE: The figure above is a sketch: the triangle is probably not actually the size shown above. That is not a problem, but it means that we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Find the value of cos(C)
[−2 points ⇒ 0 / 4 points left]

We want to find cos(C), which means we want the ratio xr. So we need the value of x. We can find this using the theorem of Pythagoras with the reference triangle:

a2+b2=c2x2+(7)2=(11)2x2=12149x2=72x2=362x2=±362x=±62

This calculation tells us two things: the third side of the triangle is 62 units long, and the x-coordinate of the point in our diagram is 62. Now we have enough information to evaluate the cosine ratio for angle C.

cos(C)=xr=6211=6211
TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant III and we expect the cosine ratio to be negative in this quadrant. So all is groovy!

The value of cos(C) is 6211.


Submit your answer as:

Finding one ratio from another

Given:

sin(C)=1921

Determine the value of tan(C) if C is obtuse.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).
Answer: tan(C)= .
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle C is in and draw a sketch to represent that angle.


STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle C points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is obtuse, the angle must be in Quadrant II. This agrees with the fact that sin(C) is positive. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the sine ratio is the ratio of y to r.

NOTE: The figure above is a sketch: the triangle is probably not actually the size shown above. That is not a problem, but it means that we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Find the value of tan(C)
[−2 points ⇒ 0 / 4 points left]

We want to find tan(C), which means we want the ratio yx. So we need the value of x. We can find this using the theorem of Pythagoras with the reference triangle:

a2+b2=c2x2+(19)2=(21)2x2=441361x2=80x2=165x2=±165x=±45

This calculation tells us two things: the third side of the triangle is 45 units long, and the x-coordinate of the point in our diagram is 45. Now we have enough information to evaluate the tangent ratio for angle C.

tan(C)=yx=1945=1945
TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant II and we expect the tangent ratio to be negative in this quadrant. So all is groovy!

The value of tan(C) is 1945.


Submit your answer as:

The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant IV, where we find the letter C:

The C is always in Quadrant IV. The cosine ratio is positive for any angle in Quadrant IV. At the same time, it tells us that the sine and tangent ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the sine ratio is positive in Quadrants I and II, while it is negative in Quadrants III and IV.

The CAST diagram tells us when the trigonometric ratios are positive or negative.


Submit your answer as:

The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant II, where we find the letter S:

The S is always in Quadrant II. The sine ratio is positive for any angle in Quadrant II. At the same time, it tells us that the cosine and tangent ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The CAST diagram tells us when the trigonometric ratios are positive or negative.


Submit your answer as:

The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.


STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

  • C - cosine
  • A - all (sine, cosine, and tangent)
  • S - sine
  • T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant IV, where we find the letter C:

The C is always in Quadrant IV. The cosine ratio is positive for any angle in Quadrant IV. At the same time, it tells us that the sine and tangent ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

  • In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between 0° and 90°.
  • Each trigonometric ratio is positive in 2 of the quadrants and negative in 2 of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The CAST diagram tells us when the trigonometric ratios are positive or negative.


Submit your answer as:

Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point E is on the circle at (16;12). The angle between the positive x-axis and Point E is e. A radius of the circle is shown from the origin to Point E.

Answer the two questions which follow about Point E and e.

  1. Determine the length of the radius, r.

    Answer: The length of the radius is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by drawing a right-angled reference triangle on the figure. This triangle should include Point E and the radius shown.


    STEP: Draw a right-angled triangle on the figure
    [−1 point ⇒ 1 / 2 points left]

    We need to find the radius of the circle. We can do this by using the coordinates of Point E. This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point E, the origin, and the x-axis, as shown below.

    Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point E. The x-coordinate of Point E is 16: that means E is 16 units away from the y-axis. Similarly, the y-coordinate of E is 12, which means that E is 12 units away from the x-axis.

    Notice that the sides of the triangle are labelled with positive numbers even though the x- and y-coordinates are negative.

    TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

    STEP: Use the triangle to calculate the radius
    [−1 point ⇒ 0 / 2 points left]

    The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

    a2+b2=c2(16)2+(12)2=r2400=r2±400=r2±20=r

    We get a radius of ±20. But it must be positive because it is a length.

    The length of the radius of the circle is 20.


    Submit your answer as:
  2. Hence determine the value of sin(e).

    Answer: The value of sin(e) is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the definition of the sine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    We need to find the value of sin(e). The trigonometric functions are defined on the Cartesian plane in terms of the coordinates (x;y) and the radius r. And for e, we know all three of those numbers from Question 1! In this case we want to know sin(e), so we need the definition of the sine ratio on the Cartesian plane:

    sin(θ)=yr

    Point E corresponds to e, so we need to substitute in the correct values from the point. In this case that means using the y-coordinate and the radius.

    y=12r=20sin(e)=1220=35
    TIP: Compare the sign of your answer to the CAST diagram. Point E is in Quadrant III. The CAST diagram tells us that the sine ratio in Quadrant III is always negative. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(e) is 35.


    Submit your answer as:

Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point G is on the circle at (6;8). The angle between the positive x-axis and Point G is g. A radius of the circle is shown from the origin to Point G.

Answer the two questions which follow about Point G and g.

  1. Determine the length of the radius, r.

    Answer: The length of the radius is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by drawing a right-angled reference triangle on the figure. This triangle should include Point G and the radius shown.


    STEP: Draw a right-angled triangle on the figure
    [−1 point ⇒ 1 / 2 points left]

    We need to find the radius of the circle. We can do this by using the coordinates of Point G. This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point G, the origin, and the x-axis, as shown below.

    Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point G. The x-coordinate of Point G is 6: that means G is 6 units away from the y-axis. Similarly, the y-coordinate of G is 8, which means that G is 8 units away from the x-axis.

    Notice that the sides of the triangle are labelled with positive numbers even though the x-coordinate is negative.

    TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

    STEP: Use the triangle to calculate the radius
    [−1 point ⇒ 0 / 2 points left]

    The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

    a2+b2=c2(6)2+(8)2=r2100=r2±100=r2±10=r

    We get a radius of ±10. But it must be positive because it is a length.

    The length of the radius of the circle is 10.


    Submit your answer as:
  2. Hence determine the value of sin(g).

    Answer: The value of sin(g) is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the definition of the sine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.


    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    We need to find the value of sin(g). The trigonometric functions are defined on the Cartesian plane in terms of the coordinates (x;y) and the radius r. And for g, we know all three of those numbers from Question 1! In this case we want to know sin(g), so we need the definition of the sine ratio on the Cartesian plane:

    sin(θ)=yr

    Point G corresponds to g, so we need to substitute in the correct values from the point. In this case that means using the y-coordinate and the radius.

    y=8r=10sin(g)=810=45
    TIP: Compare the sign of your answer to the CAST diagram. Point G is in Quadrant II. The CAST diagram tells us that the sine ratio in Quadrant II is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of sin(g) is 45.


    Submit your answer as:

Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point H is on the circle at (8;6). The angle between the positive x-axis and Point H is h. A radius of the circle is shown from the origin to Point H.

Answer the two questions which follow about Point H and h.

  1. Determine the length of the radius, r.

    Answer: The length of the radius is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by drawing a right-angled reference triangle on the figure. This triangle should include Point H and the radius shown.


    STEP: Draw a right-angled triangle on the figure
    [−1 point ⇒ 1 / 2 points left]

    We need to find the radius of the circle. We can do this by using the coordinates of Point H. This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point H, the origin, and the x-axis, as shown below.

    Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point H. The x-coordinate of Point H is 8: that means H is 8 units away from the y-axis. Similarly, the y-coordinate of H is 6, which means that H is 6 units away from the x-axis.

    Notice that the sides of the triangle are labelled with positive numbers even though the y-coordinate is negative.

    TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

    STEP: Use the triangle to calculate the radius
    [−1 point ⇒ 0 / 2 points left]

    The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

    a2+b2=c2(8)2+(6)2=r2100=r2±100=r2±10=r

    We get a radius of ±10. But it must be positive because it is a length.

    The length of the radius of the circle is 10.


    Submit your answer as:
  2. Hence determine the value of cos(h).

    Answer: The value of cos(h) is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the definition of the cosine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.


    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    We need to find the value of cos(h). The trigonometric functions are defined on the Cartesian plane in terms of the coordinates (x;y) and the radius r. And for h, we know all three of those numbers from Question 1! In this case we want to know cos(h), so we need the definition of the cosine ratio on the Cartesian plane:

    cos(θ)=xr

    Point H corresponds to h, so we need to substitute in the correct values from the point. In this case that means using the x-coordinate and the radius.

    x=8r=10cos(h)=810=45
    TIP: Compare the sign of your answer to the CAST diagram. Point H is in Quadrant IV. The CAST diagram tells us that the cosine ratio in Quadrant IV is always positive. Make sure the sign of your answer agrees with the CAST diagram.

    The value of cos(h) is 45.


    Submit your answer as:

Finding one ratio from another

Suppose cos(θ)=2029, and 180°θ360°. Without using a calculator, determine the value of sin(θ).

INSTRUCTION: Write your answer as a simplified fraction.
Answer: The value of sin(θ) is .
fraction
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which θ sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find sin(θ).


STEP: Determine which quadrant θ must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions cos(θ) and sin(θ). We know one and need to find the other. We can do this because cos(θ) tells us about two of the three values x, y, and r. We can use those to find the third value, which we need in order to find sin(θ).

The first step is to figure out which quadrant the angle θ points to. We can do this by combining the CAST diagram with the interval given in the question: 180°θ360°. We already know that cos(θ) is negative because it is equal to 2029. The CAST diagram tells us that the cosine ratio is negative in Quadrants II and III. But since θ is between 180° and 360°, it cannot be in Quadrant II.

NOTE: If we did not know that 180°θ360°, we could not know if the angle is in Quadrant II or III.

The angle θ must point to Quadrant III.


STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant III to represent angle θ.

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant III so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that cos(θ) is equal to 2029 (this is given in the question). This tells us about two sides of the triangle in the picture above.

cos(θ)=2029=xrxr=2029This means:x=20r=29

Now we know one of the coordinates of Point P. We also know two sides of the triangle. Note that the lengths of the triangle's sides must be positive because they are distances, no matter which quadrant the triangle is in.


STEP: Find the value of y
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of y. We can find this using the theorem of Pythagoras for the reference triangle in the figure.

a2+b2=c2y2+(20)2=(29)2y2=841400y2=441y=±21

For the third side of the triangle, we must take the positive value (because it is a distance). However, the sign of y depends on which quadrant the point is in. In this case, Point P is in Quadrant III, where the y-values must be negative. So the correct value for the y-coordinate is 21 (we throw away the positive answer).


STEP: Use the definition of the sine ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point P corresponds to angle θ, we can evaluate sin(θ) by substituting in the correct values from the point.

sin(θ)=yr=2129
TIP: Point P is in Quadrant III, where the CAST diagram tells us that the sine ratio is always negative. Make sure your answer agrees with that!

The value of sin(θ) is 2129.


Submit your answer as:

Finding one ratio from another

Suppose tan(ω)=158, and 180°ω360°. Without using a calculator, determine the value of sin(ω).

INSTRUCTION: Write your answer as a simplified fraction.
Answer: The value of sin(ω) is .
fraction
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which ω sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find sin(ω).


STEP: Determine which quadrant ω must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions tan(ω) and sin(ω). We know one and need to find the other. We can do this because tan(ω) tells us about two of the three values x, y, and r. We can use those to find the third value, which we need in order to find sin(ω).

The first step is to figure out which quadrant the angle ω points to. We can do this by combining the CAST diagram with the interval given in the question: 180°ω360°. We already know that tan(ω) is positive because it is equal to 158. The CAST diagram tells us that the tangent ratio is positive in Quadrants I and III. But since ω is between 180° and 360°, it cannot be in Quadrant I.

NOTE: If we did not know that 180°ω360°, we could not know if the angle is in Quadrant I or III.

The angle ω must point to Quadrant III.


STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant III to represent angle ω.

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant III so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that tan(ω) is equal to 158 (this is given in the question). This tells us about two sides of the triangle in the picture above.

tan(θ)=158=yxyx=158This means:y=15x=8

tanθ is positive. But we know that the point is in the third quadrant, which means that x and y are both negative. These negatives are hidden in the ratio because 158 is equal to 158.

Now we know both of the coordinates of Point P. We also know two sides of the triangle. Note that the lengths of the triangle's sides must be positive because they are distances, no matter which quadrant the triangle is in.


STEP: Find the value of r
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of r. We can find this using the theorem of Pythagoras for the reference triangle in the figure.

a2+b2=c2(8)2+(15)2=r264+225=r2289=r2±17=r

For the third side of the triangle, we must take the positive value (because it is a distance). The radius is always positive as well.


STEP: Use the definition of the sine ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point P corresponds to angle ω, we can evaluate sin(ω) by substituting in the correct values from the point.

sin(θ)=yr=1517
TIP: Point P is in Quadrant III, where the CAST diagram tells us that the sine ratio is always negative. Make sure your answer agrees with that!

The value of sin(ω) is 1517.


Submit your answer as:

Finding one ratio from another

Suppose cos(ω)=45, and 180°ω360°. Without using a calculator, determine the value of sin(ω).

INSTRUCTION: Write your answer as a simplified fraction.
Answer: The value of sin(ω) is .
fraction
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which ω sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find sin(ω).


STEP: Determine which quadrant ω must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions cos(ω) and sin(ω). We know one and need to find the other. We can do this because cos(ω) tells us about two of the three values x, y, and r. We can use those to find the third value, which we need in order to find sin(ω).

The first step is to figure out which quadrant the angle ω points to. We can do this by combining the CAST diagram with the interval given in the question: 180°ω360°. We already know that cos(ω) is positive because it is equal to 45. The CAST diagram tells us that the cosine ratio is positive in Quadrants I and IV. But since ω is between 180° and 360°, it cannot be in Quadrant I.

NOTE: If we did not know that 180°ω360°, we could not know if the angle is in Quadrant I or IV.

The angle ω must point to Quadrant IV.


STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant IV to represent angle ω.

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant IV so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that cos(ω) is equal to 45 (this is given in the question). This tells us about two sides of the triangle in the picture above.

cos(θ)=45=xrxr=45This means:x=4r=5

Now we know one of the coordinates of Point P. We also know two sides of the triangle. Note that the lengths of the triangle's sides must be positive because they are distances, no matter which quadrant the triangle is in.


STEP: Find the value of y
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of y. We can find this using the theorem of Pythagoras for the reference triangle in the figure.

a2+b2=c2y2+(4)2=(5)2y2=2516y2=9y=±3

For the third side of the triangle, we must take the positive value (because it is a distance). However, the sign of y depends on which quadrant the point is in. In this case, Point P is in Quadrant IV, where the y-values must be negative. So the correct value for the y-coordinate is 3 (we throw away the positive answer).


STEP: Use the definition of the sine ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point P corresponds to angle ω, we can evaluate sin(ω) by substituting in the correct values from the point.

sin(θ)=yr=35
TIP: Point P is in Quadrant IV, where the CAST diagram tells us that the sine ratio is always negative. Make sure your answer agrees with that!

The value of sin(ω) is 35.


Submit your answer as:

Working with trigonometric ratios

The diagram below shows Point P at (5;3). Triangle OPQ is also shown, with three sides labelled as 5, 3, and 34. The angle ω reaches from the positive x-axis to the line OP, as labelled. Answer the two questions which follow below.

  1. Determine the value of cosω.

    INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).
    Answer: cosω=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition for the cosine ratio on the Cartesian plane:

    cosθ=xr

    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    On the Cartesian plane the cosine ratio is defined as xr. That means that we can read the answer from the diagram in the question. We need the values for x and r for Point P.

    P is at (5;3). From the diagram we can see that the radius is r=34. So we get:

    x=5andr=34

    Putting those numbers into the definition for the cosine ratio, we get:

    cosω=xrcosω=534
    NOTE: The size of angle ω puts Point P in Quadrant IV. The CAST diagram tells us that the cosine ratio is always positive in Quadrant IV. And indeed the answer we got agrees with that: it is positive because the x-coordinate is positive (and the radius is always positive).

    The value of cosω is 534.


    Submit your answer as:
  2. Hence evaluate the following expression:

    2cos2ω23
    Answer: 2cos2ω23=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.


    STEP: Substitute the value of cosω into the expression
    [−1 point ⇒ 1 / 2 points left]

    Start by substituting the answer from Question 1 into the expression:

    2cos2ω23=2(534)223

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now evaluate the expression.

    2(534)223=2253423=251723=(253)(217)173=4151

    The value of 2cos2ω23 is 4151.


    Submit your answer as:

Working with trigonometric ratios

The diagram below shows Point P at (7;3). Triangle OPQ is also shown, with three sides labelled as 7, 3, and 58. The angle α reaches from the positive x-axis to the line OP, as labelled. Answer the two questions which follow below.

  1. Determine the value of cosα.

    INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).
    Answer: cosα=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition for the cosine ratio on the Cartesian plane:

    cosθ=xr

    STEP: Use the definition of the cosine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    On the Cartesian plane the cosine ratio is defined as xr. That means that we can read the answer from the diagram in the question. We need the values for x and r for Point P.

    P is at (7;3). From the diagram we can see that the radius is r=58. So we get:

    x=7andr=58

    Putting those numbers into the definition for the cosine ratio, we get:

    cosα=xrcosα=758
    NOTE: The size of angle α puts Point P in Quadrant IV. The CAST diagram tells us that the cosine ratio is always positive in Quadrant IV. And indeed the answer we got agrees with that: it is positive because the x-coordinate is positive (and the radius is always positive).

    The value of cosα is 758.


    Submit your answer as:
  2. Hence evaluate the following expression:

    2cos2α32
    Answer: 2cos2α32=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.


    STEP: Substitute the value of cosα into the expression
    [−1 point ⇒ 1 / 2 points left]

    Start by substituting the answer from Question 1 into the expression:

    2cos2α32=2(758)232

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now evaluate the expression.

    2(758)232=2495832=492932=(492)(329)292=1158

    The value of 2cos2α32 is 1158.


    Submit your answer as:

Working with trigonometric ratios

The diagram below shows Point P at (5;3). Triangle OPQ is also shown, with three sides labelled as 5, 3, and 34. The angle α reaches from the positive x-axis to the line OP, as labelled. Answer the two questions which follow below.

  1. Determine the value of sinα.

    INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).
    Answer: sinα=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the definition for the sine ratio on the Cartesian plane:

    sinθ=yr

    STEP: Use the definition of the sine ratio on the Cartesian plane
    [−2 points ⇒ 0 / 2 points left]

    On the Cartesian plane the sine ratio is defined as yr. That means that we can read the answer from the diagram in the question. We need the values for y and r for Point P.

    P is at (5;3). From the diagram we can see that the radius is r=34. So we get:

    y=3andr=34

    Putting those numbers into the definition for the sine ratio, we get:

    sinα=yrsinα=334
    NOTE: The size of angle α puts Point P in Quadrant III. The CAST diagram tells us that the sine ratio is always negative in Quadrant III. And indeed the answer we got agrees with that: it is negative because the y-coordinate is positive (and the radius is always positive).

    The value of sinα is 334.


    Submit your answer as:
  2. Hence evaluate the following expression:

    4sin2α52
    Answer: 4sin2α52=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.


    STEP: Substitute the value of sinα into the expression
    [−1 point ⇒ 1 / 2 points left]

    Start by substituting the answer from Question 1 into the expression:

    4sin2α52=4(334)252

    STEP: Evaluate the expression
    [−1 point ⇒ 0 / 2 points left]

    Now evaluate the expression.

    4(334)252=493452=181752=(182)(517)172=4934

    The value of 4sin2α52 is 4934.


    Submit your answer as:

Using the CAST diagram

  1. The figure below shows the CAST diagram. Point P is shown, and angle θ is the angle from the positive x-axis to Point P.

    Is sinθ positive or negative?

    Answer: sinθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Point P is in Quadrant III. What does the CAST diagram tell you about the sign of the sine ratio in that quadrant?


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to determine if the expression sinθ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

    We can see that the angle θ puts Point P in Quadrant III. On the CAST diagram, what letter is in Quadrant III? It is T. The T means that the tangent ratio is positive in that quadrant and the other ratios are negative.

    Based on the CAST diagram, the expression sinθ must be negative.


    Submit your answer as:
  2. Now determine if tanθ is positive or negative.

    Answer: tanθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the same approach as you used in Question 1.


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    For this question, we use exactly the same information that we used in Question 1. Point P and θ are in Quadrant III, where we find the letter T on the CAST diagram. This means that the tangent ratio is positive for any angle in that quadrant, while cosine and sine are negative.

    NOTE:

    What do the answers to these questions mean?

    The angle shown in the diagram is θ=220°. We can calculate the value of the expressions for these two questions (use a calculator):

    From Question 1:sin220°=0,64278...From Question 2:tan220°=0,83909...

    You can see that these values are negative and positive, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

    The expression tanθ must be positive.


    Submit your answer as:

Using the CAST diagram

  1. The figure below shows the CAST diagram. Point P is shown, and angle θ is the angle from the positive x-axis to Point P.

    Is sinθ positive or negative?

    Answer: sinθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Point P is in Quadrant I. What does the CAST diagram tell you about the sign of the sine ratio in that quadrant?


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to determine if the expression sinθ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

    We can see that the angle θ puts Point P in Quadrant I. On the CAST diagram, what letter is in Quadrant I? It is A. The A means that all three of the trigonometric ratios are positive for angles in Quadrant I.

    Based on the CAST diagram, the expression sinθ must be positive.


    Submit your answer as:
  2. Now determine if cosθ is positive or negative.

    Answer: cosθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the same approach as you used in Question 1.


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We already know the answer from Question 1: for angles in Quadrant I the trigonometric ratios are always positive. It does not even matter which ratio we are talking about! If the angle is between 0° and 90° (in Quadrant I) then the value of the ratio will be positive

    NOTE:

    What do the answers to these questions mean?

    The angle shown in the diagram is θ=43°. We can calculate the value of the expressions for these two questions (use a calculator):

    From Question 1:sin43°=0,68199...From Question 2:cos43°=0,73135...

    You can see that these values are both positive, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

    The expression cosθ must be positive.


    Submit your answer as:

Using the CAST diagram

  1. The figure below shows the CAST diagram. Point P is shown, and angle θ is the angle from the positive x-axis to Point P.

    Is sinθ positive or negative?

    Answer: sinθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Point P is in Quadrant III. What does the CAST diagram tell you about the sign of the sine ratio in that quadrant?


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    We need to determine if the expression sinθ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

    We can see that the angle θ puts Point P in Quadrant III. On the CAST diagram, what letter is in Quadrant III? It is T. The T means that the tangent ratio is positive in that quadrant and the other ratios are negative.

    Based on the CAST diagram, the expression sinθ must be negative.


    Submit your answer as:
  2. Now determine if cosθ is positive or negative.

    Answer: cosθ is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the same approach as you used in Question 1.


    STEP: Read the answer from the CAST diagram
    [−1 point ⇒ 0 / 1 points left]

    For this question, we use exactly the same information that we used in Question 1. Point P and θ are in Quadrant III, where we find the letter T on the CAST diagram. This means that the tangent ratio is positive for any angle in that quadrant, while cosine and sine are negative.

    NOTE:

    What do the answers to these questions mean?

    The angle shown in the diagram is θ=222°. We can calculate the value of the expressions for these two questions (use a calculator):

    From Question 1:sin222°=0,66913...From Question 2:cos222°=0,74314...

    You can see that these values are both negative, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

    The expression cosθ must be negative.


    Submit your answer as:

5. Trigonometric graphs

The cosine function

The table below shows input and output values for the trigonometric function f(x)=cosx. One of the output values is missing.

x f(x)
0° 1
90° 0
180° 1
270° 0
360° ?

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=cosx and we need to fill in the missing value from the table. The missing value corresponds to f(360°). So we need to evaluate the function with x=360°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(360°) is 1. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 1.


Submit your answer as:

The tangent function

The table below shows input and output values for the trigonometric function f(x)=tanx. One of the output values is missing.

x f(x)
0° 0
30° 33
60° 3
90° undefined
120° 3
150° 33
180° ?

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=tanx and we need to fill in the missing value from the table. The missing value corresponds to f(180°). So we need to evaluate the function with x=180°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(180°) is 0. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 0.


Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function f(x)=sinx. One of the output values is missing.

x f(x)
0° 0
30° 12
60° 32
90° ?
120° 32
150° 12
180° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=sinx and we need to fill in the missing value from the table. The missing value corresponds to f(90°). So we need to evaluate the function with x=90°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(90°) is 1. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 1.


Submit your answer as:

Exercises

The cosine function

The table below shows input and output values for the trigonometric function f(x)=cosx. One of the output values is missing.

x f(x)
0° 1
60° 12
120° ?
180° 1
240° 12
300° 12
360° 1

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=cosx and we need to fill in the missing value from the table. The missing value corresponds to f(120°). So we need to evaluate the function with x=120°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(120°) is 12. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 12.


Submit your answer as:

The cosine function

The table below shows input and output values for the trigonometric function f(x)=cosx. One of the output values is missing.

x f(x)
0° 1
45° 22
90° 0
135° 22
180° 1
225° 22
270° 0
315° 22
360° ?

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=cosx and we need to fill in the missing value from the table. The missing value corresponds to f(360°). So we need to evaluate the function with x=360°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(360°) is 1. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 1.


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The cosine function

The table below shows input and output values for the trigonometric function f(x)=cosx. One of the output values is missing.

x f(x)
0° ?
30° 32
60° 12
90° 0
120° 12
150° 32
180° 1

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=cosx and we need to fill in the missing value from the table. The missing value corresponds to f(0°). So we need to evaluate the function with x=0°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(0°) is 1. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 1.


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The tangent function

The table below shows input and output values for the trigonometric function f(x)=tanx. One of the output values is missing.

x f(x)
0° 0
60° ?
120° 3
180° 0
240° 3
300° 3
360° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.
Answer: The missing value is .
expression
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=tanx and we need to fill in the missing value from the table. The missing value corresponds to f(60°). So we need to evaluate the function with x=60°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(60°) is 3. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 3.


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The tangent function

The table below shows input and output values for the trigonometric function f(x)=tanx. One of the output values is missing.

x f(x)
0° 0
60° 3
120° 3
180° 0
240° 3
300° ?
360° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.
Answer: The missing value is .
expression
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=tanx and we need to fill in the missing value from the table. The missing value corresponds to f(300°). So we need to evaluate the function with x=300°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(300°) is 3. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 3.


Submit your answer as:

The tangent function

The table below shows input and output values for the trigonometric function f(x)=tanx. One of the output values is missing.

x f(x)
0° ?
90° undefined
180° 0
270° undefined
360° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=tanx and we need to fill in the missing value from the table. The missing value corresponds to f(0°). So we need to evaluate the function with x=0°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(0°) is 0. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 0.


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The sine function

The table below shows input and output values for the trigonometric function f(x)=sinx. One of the output values is missing.

x f(x)
0° 0
90° 1
180° 0
270° ?
360° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=sinx and we need to fill in the missing value from the table. The missing value corresponds to f(270°). So we need to evaluate the function with x=270°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(270°) is 1. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 1.


Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function f(x)=sinx. One of the output values is missing.

x f(x)
0° 0
45° 22
90° ?
135° 22
180° 0
225° 22
270° 1
315° 22
360° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=sinx and we need to fill in the missing value from the table. The missing value corresponds to f(90°). So we need to evaluate the function with x=90°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(90°) is 1. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 1.


Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function f(x)=sinx. One of the output values is missing.

x f(x)
0° 0
45° 22
90° 1
135° 22
180° ?
225° 22
270° 1
315° 22
360° 0

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.
Answer: The missing value is .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.


STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function f(x)=sinx and we need to fill in the missing value from the table. The missing value corresponds to f(180°). So we need to evaluate the function with x=180°.

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of f(180°) is 0. This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is 0.


Submit your answer as: